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Figure 1 shows a sketch of a curve C with equation $y = f(x)$ and a straight line l - Edexcel - A-Level Maths Pure - Question 9 - 2020 - Paper 1

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Figure 1 shows a sketch of a curve C with equation $y = f(x)$ and a straight line l. The curve C meets l at the points $(-2,13)$ and $(0,25)$ as shown. The shaded... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of a curve C with equation $y = f(x)$ and a straight line l - Edexcel - A-Level Maths Pure - Question 9 - 2020 - Paper 1

Step 1

Find the Equation of the Line l

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Answer

To find the equation of the line l that intersects the curve C at the points (2,13)(-2, 13) and (0,25)(0, 25), we can use the formula for the slope, mm:

m=y2y1x2x1=25130(2)=122=6m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{25 - 13}{0 - (-2)} = \frac{12}{2} = 6

Using the point-slope form of the equation, yy1=m(xx1)y - y_1 = m(x - x_1), we can substitute one of the points, say (0,25)(0, 25):

y25=6(x0)y=6x+25y - 25 = 6(x - 0) \Rightarrow y = 6x + 25

Step 2

Find the Equation of the Curve C

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Knowing that f(x)f(x) is a quadratic function with a minimum turning point at (2,13)(-2, 13), we express it in vertex form:

f(x)=a(x+2)2+13f(x) = a(x + 2)^2 + 13

To find the value of aa, we can use the other point (0,25)(0, 25):

25=a(0+2)2+1325=4a+134a=12a=325 = a(0 + 2)^2 + 13 \Rightarrow 25 = 4a + 13 \Rightarrow 4a = 12 \Rightarrow a = 3

Thus, the equation of the curve is:

f(x)=3(x+2)2+13f(x) = 3(x + 2)^2 + 13

Step 3

Define the Region R

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Answer

The shaded region R is defined as follows:

R={(x,y):f(x)y and y6x+25, for 2x0}R = \{(x,y) : f(x) \geq y \text{ and } y \leq 6x + 25, \text{ for } -2 \leq x \leq 0 \}

This encompasses the area between the curve C and the line l over the specified interval.

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