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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

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A curve C, has equation $y = x^2 - 4x + k$, where $k$ is a constant. It crosses the x-axis at the points $(2 + adical{5}, 0)$ and $(2 - adical{5}, 0)$. 6 (a) Find... 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 kk, we use the fact that the curve crosses the x-axis at given points. This means that the value of yy is 0 at those points. Hence, we can set up the equation:

0 &= (2 + adical{5})^2 - 4(2 + adical{5}) + k \ 0 &= (2 - adical{5})^2 - 4(2 - adical{5}) + k \end{align*}$$ Calculating $(2 + adical{5})^2$: $$(2 + adical{5})^2 = 4 + 4 adical{5} + 5 = 9 + 4 adical{5}$$ Then obtaining the value: $$egin{align*} 9 + 4 adical{5} - 8 - 4 adical{5} + k &= 0 \ k &= -1 \end{align*}$$ So the value of $k$ is $-1$.

Step 2

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

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Answer

The curve C is defined by the equation:

y=x24x1y = x^2 - 4x - 1

To sktech the graph:

  1. Notice that the x-intercepts are (2+adical5,0)(2 + adical{5}, 0) and (2adical5,0)(2 - adical{5}, 0).
  2. The vertex of the parabola is at the point (2,5)(2, -5), found using the formula for the vertex of a parabola x = - rac{b}{2a}.
  3. The y-intercept can be found by setting x=0x = 0:

y=024(0)1=1y = 0^2 - 4(0) - 1 = -1

Thus, the curve intersects the y-axis at (0,1)(0, -1).

The graph should be a U-shaped curve that opens upwards, intersecting the x-axis at the calculated points, with the vertex below the x-axis and passing through the y-axis at (0,1)(0, -1). Be sure to accurately label all intersections on the graph.

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