💡 GCSE Maths (Foundation): Quadratics and Graphing Practice Questions
1
Solved Example
Easy Level
Question 1: Identify which of the following graphs represents a quadratic function. 🤔
Graph A: A straight line passing through the origin.
Graph B: A U-shaped curve that opens upwards, passing through \((-2, 4)\), \((0, 0)\), and \((2, 4)\).
Graph C: A curve that starts low, goes up, then comes back down, like an 'S' shape.
Solution & Explanation
Here's how to identify the quadratic graph:
Step 1: Understand Quadratic Functions 💡 A quadratic function is typically of the form \(y = ax^2 + bx + c\), where \(a \neq 0\). Its graph is always a U-shaped or an inverted U-shaped curve called a parabola.
Step 2: Analyze Graph A Graph A is a straight line. Straight lines represent linear functions (e.g., \(y = mx + c\)), not quadratic functions.
Step 3: Analyze Graph B ✅ Graph B is described as a U-shaped curve that opens upwards. This is the characteristic shape of a parabola, which is the graph of a quadratic function (e.g., \(y=x^2\)).
Step 4: Analyze Graph C Graph C is an 'S' shape curve. This typically represents a cubic function or another type of polynomial, not a quadratic function.
Conclusion: 📈 Graph B represents a quadratic function because it is a parabola.
2
Solved Example
Medium Level
Question 2: Complete the table of values for the quadratic function \(y = x^2 - 2\) and then state the coordinates of the turning point.
| \(x\) | -2 | -1 | 0 | 1 | 2 |
| \(y\) | | | | | |
Solution & Explanation
Let's complete the table and find the turning point:
Step 1: Calculate \(y\) for each \(x\) value 👉
For \(x = -2\): \(y = (-2)^2 - 2 = 4 - 2 = 2\)
For \(x = -1\): \(y = (-1)^2 - 2 = 1 - 2 = -1\)
For \(x = 0\): \(y = (0)^2 - 2 = 0 - 2 = -2\)
For \(x = 1\): \(y = (1)^2 - 2 = 1 - 2 = -1\)
For \(x = 2\): \(y = (2)^2 - 2 = 4 - 2 = 2\)
Step 2: Complete the table ✅
| \(x\) | -2 | -1 | 0 | 1 | 2 |
| \(y\) | 2 | -1 | -2 | -1 | 2 |
Step 3: Identify the turning point 📌 The turning point (also called the vertex) is the point where the graph changes direction. Looking at the \(y\) values \((2, -1, -2, -1, 2)\), the lowest \(y\) value is \(-2\), which occurs when \(x = 0\).
Conclusion: The completed table is shown above. The coordinates of the turning point are \((0, -2)\).
3
Solved Example
Easy Level
Question 3: A quadratic graph crosses the x-axis at \((-3, 0)\) and \((1, 0)\). It crosses the y-axis at \((0, -3)\). State the x-intercepts and the y-intercept of the graph.
Solution & Explanation
This question asks us to identify intercepts from given points. 💡
Step 1: Understand Intercepts
An x-intercept is a point where the graph crosses or touches the x-axis. At these points, the \(y\)-coordinate is always 0.
A y-intercept is a point where the graph crosses or touches the y-axis. At these points, the \(x\)-coordinate is always 0.
Step 2: Identify x-intercepts
The problem states the graph crosses the x-axis at \((-3, 0)\) and \((1, 0)\). These are the points where \(y=0\).
Step 3: Identify y-intercept
The problem states the graph crosses the y-axis at \((0, -3)\). This is the point where \(x=0\).
Conclusion:
The x-intercepts are \((-3, 0)\) and \((1, 0)\).
The y-intercept is \((0, -3)\).
4
Solved Example
Medium Level
Question 4: Consider the graph of \(y = x^2 - 4x + 3\). By observing the graph (imagine a parabola opening upwards, passing through \((0, 3)\), \((1, 0)\), \((2, -1)\), \((3, 0)\), \((4, 3)\)), identify its turning point and the equation of its axis of symmetry. 📌
Solution & Explanation
Let's find the turning point and axis of symmetry from the described graph.
Step 1: Locate the Turning Point 📈
The turning point is the lowest (minimum) or highest (maximum) point on a parabola. For a parabola that opens upwards (like \(y=x^2-4x+3\)), the turning point is the minimum point. From the given points, the lowest \(y\)-value is \(-1\), which occurs at \((2, -1)\). Therefore, the turning point is \((2, -1)\).
Step 2: Determine the Axis of Symmetry 💡
The axis of symmetry is a vertical line that passes through the turning point of the parabola, dividing it into two symmetrical halves. The equation of this line is always \(x = \text{the x-coordinate of the turning point}\).
Step 3: Write the Equation of the Axis of Symmetry ✅
Since the turning point is \((2, -1)\), the x-coordinate of the turning point is \(2\).
Conclusion:
The turning point is \((2, -1)\).
The equation of the axis of symmetry is \(x = 2\).
5
Solved Example
Medium Level
Question 5: Sketch the graph of \(y = -x^2 + 4\). Clearly indicate the shape of the graph, its y-intercept, and its turning point. 🤔
Solution & Explanation
Let's sketch this quadratic graph step-by-step.
Step 1: Determine the Shape of the Parabola 💡
The coefficient of \(x^2\) is \(-1\). Because it's negative, the parabola will be an inverted U-shape (opens downwards).
Step 2: Find the y-intercept 📌
The y-intercept occurs when \(x = 0\).
\(y = -(0)^2 + 4 = 0 + 4 = 4\).
So, the y-intercept is \((0, 4)\).
Step 3: Find the Turning Point 📈
For a quadratic in the form \(y = ax^2 + c\), the turning point is at \((0, c)\). In this case, \(c = 4\). This is also the y-intercept.
Alternatively, since the parabola opens downwards, the turning point will be the maximum point. As \(x^2\) is always non-negative, \(-x^2\) is always non-positive. The maximum value of \(-x^2\) is 0 (when \(x=0\)). So, the maximum value of \(y\) is \(0 + 4 = 4\), occurring at \(x=0\).
The turning point is \((0, 4)\).
Step 4: Find the x-intercepts (optional but helpful for sketching)
The x-intercepts occur when \(y = 0\).
\(0 = -x^2 + 4\)
\(x^2 = 4\)
\(x = \pm\sqrt{4}\)
\(x = 2\) or \(x = -2\)
So, the x-intercepts are \((-2, 0)\) and \((2, 0)\).
Step 5: Sketch the Graph ✅
Imagine plotting the points \((-2, 0)\), \((0, 4)\), and \((2, 0)\). Draw a smooth, inverted U-shaped curve passing through these points, with its peak (turning point) at \((0, 4)\).
Conclusion: The graph is an inverted U-shape. Its y-intercept is \((0, 4)\), which is also its turning point (maximum point).
6
Solved Example
Real World Example
Question 6: The height \(h\) (in metres) of a ball thrown upwards after \(t\) seconds is modelled by the quadratic equation \(h = 5t - t^2\). A graph of this function shows a parabola opening downwards. Using the graph, it is observed that the ball reaches its maximum height at \(t = 2.5\) seconds and lands back on the ground at \(t = 5\) seconds.
a) What is the maximum height the ball reaches? 🌍
b) What is the height of the ball after 1 second?
Solution & Explanation
Let's interpret the real-world scenario using the quadratic model.
Step 1: Understand the Model 💡
The equation \(h = 5t - t^2\) describes the height of the ball over time. The graph of this function is a parabola.
Part a) Find the Maximum Height 📈
The problem states that the ball reaches its maximum height at \(t = 2.5\) seconds. To find this maximum height, we substitute \(t = 2.5\) into the equation:
\(h = 5(2.5) - (2.5)^2\)
\(h = 12.5 - 6.25\)
\(h = 6.25\) metres.
Part b) Find the Height after 1 Second 👉
To find the height after 1 second, substitute \(t = 1\) into the equation:
\(h = 5(1) - (1)^2\)
\(h = 5 - 1\)
\(h = 4\) metres.
Conclusion:
a) The maximum height the ball reaches is \(6.25\) metres.
b) The height of the ball after 1 second is \(4\) metres.
7
Solved Example
Medium Level
Question 7: Describe the transformation that maps the graph of \(y = x^2\) onto the graph of \(y = x^2 + 3\). How does this affect the turning point?
Solution & Explanation
Let's analyze the transformation between these two quadratic functions. 💡
Step 1: Compare the Equations
We are comparing \(y = x^2\) with \(y = x^2 + 3\). The only difference is the addition of a constant, \(+3\), to the original function.
Step 2: Understand the Effect of Adding a Constant 📌
When a constant is added to a function, it results in a vertical translation (shift) of the entire graph. If the constant is positive, the graph shifts upwards. If it's negative, it shifts downwards.
Step 3: Describe the Transformation ✅
Since \(3\) is added to \(x^2\), the graph of \(y = x^2\) is shifted upwards by 3 units.
Step 4: Analyze the Effect on the Turning Point 📈
The turning point of \(y = x^2\) is at \((0, 0)\).
For \(y = x^2 + 3\), when \(x=0\), \(y = 0^2 + 3 = 3\). The lowest value of \(x^2\) is 0, so the lowest value of \(y = x^2 + 3\) is \(0+3=3\).
Thus, the turning point of \(y = x^2 + 3\) is at \((0, 3)\). The x-coordinate remains the same, but the y-coordinate increases by 3.
Conclusion: The graph of \(y = x^2\) is transformed onto the graph of \(y = x^2 + 3\) by a vertical translation of 3 units upwards. The turning point shifts from \((0, 0)\) to \((0, 3)\).
8
Solved Example
Easy Level
Question 8: The graph of \(y = x^2 - 9\) is shown below (imagine a parabola opening upwards, with its turning point at \((0, -9)\) and crossing the x-axis at \((-3, 0)\) and \((3, 0)\)). Use this graph to find the solutions to the equation \(x^2 - 9 = 0\).
Solution & Explanation
Let's find the solutions to the equation using the provided graph information. 💡
Step 1: Understand Solutions from a Graph 📌
The solutions (or roots) of a quadratic equation \(ax^2 + bx + c = 0\) are the \(x\)-values where the graph of \(y = ax^2 + bx + c\) crosses or touches the x-axis. These are also known as the x-intercepts.
Step 2: Identify the x-intercepts from the Graph Description 📈
The problem states that the graph of \(y = x^2 - 9\) crosses the x-axis at \((-3, 0)\) and \((3, 0)\).
Step 3: State the Solutions ✅
The \(x\)-coordinates of these x-intercepts are the solutions to \(x^2 - 9 = 0\).
Conclusion: The solutions to the equation \(x^2 - 9 = 0\) are \(x = -3\) and \(x = 3\).
GCSE Maths (Foundation): Quadratics and Graphing Practice Questions
Example 1:
Question 1: Identify which of the following graphs represents a quadratic function. 🤔
Graph A: A straight line passing through the origin.
Graph B: A U-shaped curve that opens upwards, passing through \((-2, 4)\), \((0, 0)\), and \((2, 4)\).
Graph C: A curve that starts low, goes up, then comes back down, like an 'S' shape.
Solution:
Here's how to identify the quadratic graph:
Step 1: Understand Quadratic Functions 💡 A quadratic function is typically of the form \(y = ax^2 + bx + c\), where \(a \neq 0\). Its graph is always a U-shaped or an inverted U-shaped curve called a parabola.
Step 2: Analyze Graph A Graph A is a straight line. Straight lines represent linear functions (e.g., \(y = mx + c\)), not quadratic functions.
Step 3: Analyze Graph B ✅ Graph B is described as a U-shaped curve that opens upwards. This is the characteristic shape of a parabola, which is the graph of a quadratic function (e.g., \(y=x^2\)).
Step 4: Analyze Graph C Graph C is an 'S' shape curve. This typically represents a cubic function or another type of polynomial, not a quadratic function.
Conclusion: 📈 Graph B represents a quadratic function because it is a parabola.
Example 2:
Question 2: Complete the table of values for the quadratic function \(y = x^2 - 2\) and then state the coordinates of the turning point.
| \(x\) | -2 | -1 | 0 | 1 | 2 |
| \(y\) | | | | | |
Solution:
Let's complete the table and find the turning point:
Step 1: Calculate \(y\) for each \(x\) value 👉
For \(x = -2\): \(y = (-2)^2 - 2 = 4 - 2 = 2\)
For \(x = -1\): \(y = (-1)^2 - 2 = 1 - 2 = -1\)
For \(x = 0\): \(y = (0)^2 - 2 = 0 - 2 = -2\)
For \(x = 1\): \(y = (1)^2 - 2 = 1 - 2 = -1\)
For \(x = 2\): \(y = (2)^2 - 2 = 4 - 2 = 2\)
Step 2: Complete the table ✅
| \(x\) | -2 | -1 | 0 | 1 | 2 |
| \(y\) | 2 | -1 | -2 | -1 | 2 |
Step 3: Identify the turning point 📌 The turning point (also called the vertex) is the point where the graph changes direction. Looking at the \(y\) values \((2, -1, -2, -1, 2)\), the lowest \(y\) value is \(-2\), which occurs when \(x = 0\).
Conclusion: The completed table is shown above. The coordinates of the turning point are \((0, -2)\).
Example 3:
Question 3: A quadratic graph crosses the x-axis at \((-3, 0)\) and \((1, 0)\). It crosses the y-axis at \((0, -3)\). State the x-intercepts and the y-intercept of the graph.
Solution:
This question asks us to identify intercepts from given points. 💡
Step 1: Understand Intercepts
An x-intercept is a point where the graph crosses or touches the x-axis. At these points, the \(y\)-coordinate is always 0.
A y-intercept is a point where the graph crosses or touches the y-axis. At these points, the \(x\)-coordinate is always 0.
Step 2: Identify x-intercepts
The problem states the graph crosses the x-axis at \((-3, 0)\) and \((1, 0)\). These are the points where \(y=0\).
Step 3: Identify y-intercept
The problem states the graph crosses the y-axis at \((0, -3)\). This is the point where \(x=0\).
Conclusion:
The x-intercepts are \((-3, 0)\) and \((1, 0)\).
The y-intercept is \((0, -3)\).
Example 4:
Question 4: Consider the graph of \(y = x^2 - 4x + 3\). By observing the graph (imagine a parabola opening upwards, passing through \((0, 3)\), \((1, 0)\), \((2, -1)\), \((3, 0)\), \((4, 3)\)), identify its turning point and the equation of its axis of symmetry. 📌
Solution:
Let's find the turning point and axis of symmetry from the described graph.
Step 1: Locate the Turning Point 📈
The turning point is the lowest (minimum) or highest (maximum) point on a parabola. For a parabola that opens upwards (like \(y=x^2-4x+3\)), the turning point is the minimum point. From the given points, the lowest \(y\)-value is \(-1\), which occurs at \((2, -1)\). Therefore, the turning point is \((2, -1)\).
Step 2: Determine the Axis of Symmetry 💡
The axis of symmetry is a vertical line that passes through the turning point of the parabola, dividing it into two symmetrical halves. The equation of this line is always \(x = \text{the x-coordinate of the turning point}\).
Step 3: Write the Equation of the Axis of Symmetry ✅
Since the turning point is \((2, -1)\), the x-coordinate of the turning point is \(2\).
Conclusion:
The turning point is \((2, -1)\).
The equation of the axis of symmetry is \(x = 2\).
Example 5:
Question 5: Sketch the graph of \(y = -x^2 + 4\). Clearly indicate the shape of the graph, its y-intercept, and its turning point. 🤔
Solution:
Let's sketch this quadratic graph step-by-step.
Step 1: Determine the Shape of the Parabola 💡
The coefficient of \(x^2\) is \(-1\). Because it's negative, the parabola will be an inverted U-shape (opens downwards).
Step 2: Find the y-intercept 📌
The y-intercept occurs when \(x = 0\).
\(y = -(0)^2 + 4 = 0 + 4 = 4\).
So, the y-intercept is \((0, 4)\).
Step 3: Find the Turning Point 📈
For a quadratic in the form \(y = ax^2 + c\), the turning point is at \((0, c)\). In this case, \(c = 4\). This is also the y-intercept.
Alternatively, since the parabola opens downwards, the turning point will be the maximum point. As \(x^2\) is always non-negative, \(-x^2\) is always non-positive. The maximum value of \(-x^2\) is 0 (when \(x=0\)). So, the maximum value of \(y\) is \(0 + 4 = 4\), occurring at \(x=0\).
The turning point is \((0, 4)\).
Step 4: Find the x-intercepts (optional but helpful for sketching)
The x-intercepts occur when \(y = 0\).
\(0 = -x^2 + 4\)
\(x^2 = 4\)
\(x = \pm\sqrt{4}\)
\(x = 2\) or \(x = -2\)
So, the x-intercepts are \((-2, 0)\) and \((2, 0)\).
Step 5: Sketch the Graph ✅
Imagine plotting the points \((-2, 0)\), \((0, 4)\), and \((2, 0)\). Draw a smooth, inverted U-shaped curve passing through these points, with its peak (turning point) at \((0, 4)\).
Conclusion: The graph is an inverted U-shape. Its y-intercept is \((0, 4)\), which is also its turning point (maximum point).
Example 6:
Question 6: The height \(h\) (in metres) of a ball thrown upwards after \(t\) seconds is modelled by the quadratic equation \(h = 5t - t^2\). A graph of this function shows a parabola opening downwards. Using the graph, it is observed that the ball reaches its maximum height at \(t = 2.5\) seconds and lands back on the ground at \(t = 5\) seconds.
a) What is the maximum height the ball reaches? 🌍
b) What is the height of the ball after 1 second?
Solution:
Let's interpret the real-world scenario using the quadratic model.
Step 1: Understand the Model 💡
The equation \(h = 5t - t^2\) describes the height of the ball over time. The graph of this function is a parabola.
Part a) Find the Maximum Height 📈
The problem states that the ball reaches its maximum height at \(t = 2.5\) seconds. To find this maximum height, we substitute \(t = 2.5\) into the equation:
\(h = 5(2.5) - (2.5)^2\)
\(h = 12.5 - 6.25\)
\(h = 6.25\) metres.
Part b) Find the Height after 1 Second 👉
To find the height after 1 second, substitute \(t = 1\) into the equation:
\(h = 5(1) - (1)^2\)
\(h = 5 - 1\)
\(h = 4\) metres.
Conclusion:
a) The maximum height the ball reaches is \(6.25\) metres.
b) The height of the ball after 1 second is \(4\) metres.
Example 7:
Question 7: Describe the transformation that maps the graph of \(y = x^2\) onto the graph of \(y = x^2 + 3\). How does this affect the turning point?
Solution:
Let's analyze the transformation between these two quadratic functions. 💡
Step 1: Compare the Equations
We are comparing \(y = x^2\) with \(y = x^2 + 3\). The only difference is the addition of a constant, \(+3\), to the original function.
Step 2: Understand the Effect of Adding a Constant 📌
When a constant is added to a function, it results in a vertical translation (shift) of the entire graph. If the constant is positive, the graph shifts upwards. If it's negative, it shifts downwards.
Step 3: Describe the Transformation ✅
Since \(3\) is added to \(x^2\), the graph of \(y = x^2\) is shifted upwards by 3 units.
Step 4: Analyze the Effect on the Turning Point 📈
The turning point of \(y = x^2\) is at \((0, 0)\).
For \(y = x^2 + 3\), when \(x=0\), \(y = 0^2 + 3 = 3\). The lowest value of \(x^2\) is 0, so the lowest value of \(y = x^2 + 3\) is \(0+3=3\).
Thus, the turning point of \(y = x^2 + 3\) is at \((0, 3)\). The x-coordinate remains the same, but the y-coordinate increases by 3.
Conclusion: The graph of \(y = x^2\) is transformed onto the graph of \(y = x^2 + 3\) by a vertical translation of 3 units upwards. The turning point shifts from \((0, 0)\) to \((0, 3)\).
Example 8:
Question 8: The graph of \(y = x^2 - 9\) is shown below (imagine a parabola opening upwards, with its turning point at \((0, -9)\) and crossing the x-axis at \((-3, 0)\) and \((3, 0)\)). Use this graph to find the solutions to the equation \(x^2 - 9 = 0\).
Solution:
Let's find the solutions to the equation using the provided graph information. 💡
Step 1: Understand Solutions from a Graph 📌
The solutions (or roots) of a quadratic equation \(ax^2 + bx + c = 0\) are the \(x\)-values where the graph of \(y = ax^2 + bx + c\) crosses or touches the x-axis. These are also known as the x-intercepts.
Step 2: Identify the x-intercepts from the Graph Description 📈
The problem states that the graph of \(y = x^2 - 9\) crosses the x-axis at \((-3, 0)\) and \((3, 0)\).
Step 3: State the Solutions ✅
The \(x\)-coordinates of these x-intercepts are the solutions to \(x^2 - 9 = 0\).
Conclusion: The solutions to the equation \(x^2 - 9 = 0\) are \(x = -3\) and \(x = 3\).