Figure 1 shows the sketch of a curve with equation $y = f(x)$, $x ext{ } ext{ } ext{} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } \ ext{The curve crosses the } y ext{-axis at } (0, 4) ext{ and crosses the } x ext{-axis at } (5, 0). \ ext{The curve has a single turning point, a maximum, at } (2, 7). \ ext{The line with equation } y = 1 ext{ is the only asymptote to the curve.} (a) State the coordinates of the turning point on the curve with equation $y = f(x - 2)$. (b) State the solution of the equation $f(2x) = 0$. (c) State the equation of the asymptote to the curve with equation $y = f(-x)$. (d) Given that the line with equation $y = k$, where $k$ is a constant, meets the curve $y = f(x)$ at only one point, state the set of possible values for $k.$

A-Level Edexcel Maths Pure Graphs of Functions: Figure 1 shows the sketch of a c

Figure 1 shows the sketch of a curve with equation $y = f(x)$, $x ext{ } ext{ } ext{} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } \ ext{The curve crosses the } y ext{-axis at } (0, 4) ext{ and crosses the } x ext{-axis at } (5, 0) - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 1

Question 8

$Figure-1-shows-the-sketch-of-a-curve-with-equation-$y-=-f(x)$,-$x--ext{-}--ext{-}--ext{}--ext{-}--ext{-}-ext{-}--ext{-}---ext{-}--ext{-}-ext{-}--ext{-}--ext{-}-ext{-}--ext{-}--ext{-}--ext{-}-ext{-}--ext{-}--ext{-}--\--ext{The-curve-crosses-the-}-y-ext{-axis-at-}-(0,-4)--ext{-and-crosses-the-}-x-ext{-axis-at-}-(5,-0)-Edexcel-A-Level Maths Pure-Question 8-2018-Paper 1.png$

Figure 1 shows the sketch of a curve with equation $y = f(x)$, $x ext{ } ext{ } ext{} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ ... show full transcript

Worked Solution & Example Answer:Figure 1 shows the sketch of a curve with equation $y = f(x)$, $x ext{ } ext{ } ext{} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } \ ext{The curve crosses the } y ext{-axis at } (0, 4) ext{ and crosses the } x ext{-axis at } (5, 0) - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 1

Step 1

State the coordinates of the turning point on the curve with equation $y = f(x - 2)$.

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Answer

The coordinates of the turning point for the transformed curve $y = f(x - 2)$ can be found by shifting the original turning point $(2, 7)$ to the right by 2 units. Therefore, the new coordinates are $(2 + 2, 7) = (4, 7)$ .

Step 2

State the solution of the equation $f(2x) = 0$.

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Answer

To solve the equation $f(2x) = 0$ , we need to find the value of $x$ that gives an output of 0 in the original function $f(x)$ . The original function crosses the $x$ -axis at $x = 5$ . Therefore, setting $2x = 5$ gives us:

$2x = 5 \ ext{ } \ x = rac{5}{2} = 2.5$

Thus, the solution is $x = 2.5$ .

Step 3

State the equation of the asymptote to the curve with equation $y = f(-x)$.

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Answer

The original asymptote is given as $y = 1$ . In this case, reflecting the curve about the $y$ -axis does not change the position of the asymptote. Therefore, the equation of the asymptote of the curve $y = f(-x)$ remains:

$y = 1$ .

Step 4

State the set of possible values for $k$.

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Answer

For the line $y = k$ to meet the curve $y = f(x)$ at only one point, $k$ can be equal to the maximum $y$ -value of the curve or less than the horizontal asymptote. Given the maximum value at the turning point $(2, 7)$ and the asymptote at $y = 1$ , the set of possible values for $k$ is:

$k ext{ } < 1 ext{ } ext{ or } k = 7$ .

State the coordinates of the turning point on the curve with equation $y = f(x - 2)$.

State the solution of the equation $f(2x) = 0$.

State the equation of the asymptote to the curve with equation $y = f(-x)$.

State the set of possible values for $k$.

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