A curve C, has equation $y = x^2 - 4x + k$, where $k$ is a constant. It crosses the x-axis at the points $(2 + oot{5}, 0)$ and $(2 - oot{5}, 0)$. 6 (a) Find the value of $k$. 6 (b) Sketch the curve C, labelling the exact values of all intersections with the axes.

A-Level AQA Maths Pure Graphs of Functions: A curve C, has equation $y = x^2

A curve C, has equation $y = x^2 - 4x + k$, where $k$ is a constant - AQA - A-Level Maths Pure - Question 6 - 2017 - Paper 2

Question 6

A curve C, has equation $y = x^2 - 4x + k$, where $k$ is a constant. It crosses the x-axis at the points $(2 + oot{5}, 0)$ and $(2 - oot{5}, 0)$. 6 (a) Find the v... show full transcript

Worked Solution & Example Answer:A curve C, has equation $y = x^2 - 4x + k$, where $k$ is a constant - AQA - A-Level Maths Pure - Question 6 - 2017 - Paper 2

Step 1

Find the value of k.

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Answer

To find the value of $k$ , we utilize the fact that the curve crosses the x-axis at the given points. This means that at these points, $y = 0$ .

We can use the factored form of the quadratic equation, where the roots are $(2 + oot{5})$ and $(2 - oot{5})$ .

The general form of a quadratic that crosses the x-axis at two points can be expressed as:

oot{5}))(x - (2 - oot{5}))$$ Expanding this expression: $$y = a((x - (2 + oot{5}))(x - (2 - oot{5})))$$ The expression becomes: $$y = a((x - 2 - oot{5})(x - 2 + oot{5}))$$ This simplifies to: $$y = a((x - 2)^2 - ( oot{5})^2)$$ Substituting $ oot{5}^2 = 5$ gives: $$y = a((x - 2)^2 - 5) = a(x^2 - 4x + 4 - 5) = a(x^2 - 4x - 1)$$ In order for this to match the original equation $y = x^2 - 4x + k$, we need to find: $$a = 1, k = -1$$ Thus: $$k = -1$$

Step 2

Sketch the curve C, labelling the exact values of all intersections with the axes.

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Answer

To sketch the curve, we start by noting the intersections with the x-axis, which we found in part (a) to be at the points $(2 + oot{5}, 0)$ and $(2 - oot{5}, 0)$ . The value of $k = -1$ indicates that the curve will intersect the y-axis at $(0, -1)$ .

Steps to Sketch:

Plot the x-intercepts: Mark points $(2 + oot{5}, 0)$ and $(2 - oot{5}, 0)$ on the x-axis.
Plot the y-intercept: Mark the point $(0, -1)$ on the y-axis.
Shape of the parabola: Since the coefficient of $x^2$ is positive (as it is a standard upward facing parabola), draw a curve that opens upwards, passing through the x-axis at the plotted points and the y-axis at $(0, -1)$ .
Label the intersections: Clearly label the points of intersection with the axes in your sketch for clarity.

Final Sketch:

Make sure the graph is of the correct shape with a vertex somewhere between the x-intercepts and smoothly curves upwards.