Comprehensive assessment covering all 6 lessons: translations, reflections, and dilations of functions, the effect of $y = f(x) + k$, $y = f(x - h)$, $y = af(x)$, and $y = f(bx)$, combined transformations, sketching and modelling, and and circles written in centre-radius and general form.
Assessment
Select the best answer for each question. 1 mark each.
The graph of $y = f(x) - 4$ is obtained from the graph of $y = f(x)$ by a translation of:
The graph of $y = f(x + 5)$ is obtained from the graph of $y = f(x)$ by a translation of:
The point $(2, -3)$ lies on the graph of $y = f(x)$. Which point lies on the graph of $y = f(x) - 2$?
The graph of $y = f(-x)$ is obtained from the graph of $y = f(x)$ by a reflection in the:
The point $(-4, 6)$ lies on the graph of $y = f(x)$. Which point lies on the graph of $y = -f(x)$?
The graph of $y = 5f(x)$ is obtained from the graph of $y = f(x)$ by a:
The graph of $y = f(4x)$ is obtained from the graph of $y = f(x)$ by a:
The point $(8, 3)$ lies on the graph of $y = f(x)$. Which point lies on the graph of $y = f\left(\frac{x}{4}\right)$?
The vertex of $y = x^2$ is at $(0, 0)$. What is the vertex of $y = 2(x + 1)^2 - 3$?
The point $(3, 2)$ lies on the graph of $y = f(x)$. Which point lies on the graph of $y = -f(x - 1) + 4$?
The graph of $y = -2f(x)$ involves which two transformations?
The graph of $f(x) = x^2$ is reflected in the $x$-axis, then vertically dilated by factor $3$, then translated $2$ units to the right. The equation of the new graph is:
The graph of $y = f(x)$ has $x$-intercepts at $x = 2$ and $x = -6$. What are the $x$-intercepts of $y = f(-x)$?
What are the centre and radius of the circle $(x + 3)^2 + (y - 1)^2 = 16$?
A ball is thrown so that its path is a parabola starting at ground level, reaching a maximum height of $5$ m after travelling $4$ m horizontally. Taking $x$ as the horizontal distance and $y$ as the height, which equation could model the path?
Short Answer
Describe fully the sequence of transformations that maps $y = f(x)$ to $y = f(x - 3) + 2$, and state the image of the point $(1, 5)$ under this transformation. Show all working.
The graph of $y = f(x)$ passes through the points $(-2, 3)$ and $(4, -1)$. State the coordinates of the images of these two points on the graph of $y = f(-x)$. Show all working.
Starting from $y = x^2$, write the equation of the parabola after a vertical dilation by factor $3$, a reflection in the $x$-axis, and a translation $4$ units up. State the coordinates of its vertex. Show all working.
The graph of $y = f(x)$ has a turning point at $(6, -2)$. Describe the transformations represented by $y = f\left(\frac{x}{2}\right)$ and hence find the coordinates of the corresponding turning point. Show all working.
Describe fully the sequence of transformations, in order, that maps $y = f(x)$ to $y = -3f(x + 1) - 4$. Show all working.
The point $(4, 6)$ lies on the graph of $y = f(x)$. Find the image of this point on the graph of $y = 2f(x - 3) + 1$, applying each transformation in turn and showing the intermediate coordinates. Show all working.
The parabola $y = x^2$ is transformed so that its new equation is $y = -2(x - 3)^2 + 8$. (a) Describe the sequence of transformations applied to $y = x^2$. (b) State the coordinates of the vertex and whether it is a maximum or a minimum. (c) Find the $y$-intercept of the transformed parabola. Show all working.
The circle $x^2 + y^2 + 6x - 8y + 9 = 0$ is given in general form. By completing the square, find its centre and radius, and hence state its domain and range. Show all working.
Q1: B, subtracting a constant outside the function, $f(x) - 4$, shifts the graph vertically. Since the constant is negative, the shift is 4 units down.
Q2: A, replacing $x$ with $x + 5$ inside the function shifts the graph horizontally. For $f(x + 5)$, the shift is 5 units to the left.
Q3: D, a vertical translation of 2 units down affects only the $y$-coordinate: $(2, -3) \to (2, -3 - 2) = (2, -5)$.
Q4: B, replacing $x$ with $-x$ inside the function reflects the graph in the $y$-axis, mapping $(x, y)$ to $(-x, y)$.
Q5: C, $-f(x)$ reflects in the $x$-axis, which negates the $y$-coordinate: $(-4, 6) \to (-4, -6)$.
Q6: D, a coefficient outside the function, $af(x)$, gives a vertical dilation. Since $a = 5$, it is a vertical dilation by factor 5.
Q7: A, a coefficient inside the function, $f(bx)$, gives a horizontal dilation by factor $\frac{1}{b}$. Since $b = 4$, the factor is $\frac{1}{4}$.
Q8: C, $f\left(\frac{x}{4}\right)$ is a horizontal dilation by factor 4. The $y$-coordinate is unchanged and the $x$-coordinate is multiplied by 4: $(8, 3) \to (32, 3)$.
Q9: D, $(x + 1)^2$ shifts left 1, and $-3$ shifts down 3. The vertical dilation by 2 does not move the vertex. So $(0, 0) \to (-1, -3)$.
Q10: B, right 1: $(3, 2) \to (4, 2)$; reflect in $x$-axis (negate $y$): $(4, -2)$; up 4: $(4, -2 + 4) = (4, 2)$.
Q11: A, the coefficient $-2$ is outside the function. Its magnitude gives a vertical dilation by factor 2, and the negative sign gives a reflection in the $x$-axis.
Q12: C, vertical dilation by 3 then reflection in the $x$-axis gives $-3x^2$; translating 2 units right replaces $x$ with $x - 2$, giving $y = -3(x - 2)^2$.
Q13: D, $f(-x)$ reflects in the $y$-axis, negating each $x$-intercept. So $x = 2 \to x = -2$ and $x = -6 \to x = 6$.
Q14: A, reverse the sign inside each bracket to read the centre $(-3, 1)$, and take the square root of the right-hand side, since it is $r^2$, to get radius 4.
Q15: D, a downward parabola (maximum) with vertex at $(4, 5)$ is $y = -(x - 4)^2 + 5$. Checking $x = 0$: $y = -16 + 5 = -11$, and $y = 4$ (positive) near the launch as required, and the maximum is 5 m at $x = 4$.
Q16 (2 marks): $f(x - 3)$ is a translation 3 units to the right, and $+2$ is a translation 2 units up [1]. Applying both to $(1, 5)$: $x \to 1 + 3 = 4$ and $y \to 5 + 2 = 7$, so the image is $(4, 7)$ [1].
Q17 (2 marks): $y = f(-x)$ is a reflection in the $y$-axis, which negates the $x$-coordinate and leaves $y$ unchanged [1]. So $(-2, 3) \to (2, 3)$ and $(4, -1) \to (-4, -1)$ [1].
Q18 (3 marks): Vertical dilation by factor 3: $y = 3x^2$ [1]. Reflection in the $x$-axis: $y = -3x^2$ [1]. Translation 4 units up: $y = -3x^2 + 4$, with vertex at $(0, 4)$ [1].
Q19 (3 marks): $f\left(\frac{x}{2}\right)$ is a horizontal dilation by factor 2 from the $y$-axis [1]. The $y$-coordinate is unchanged and the $x$-coordinate is multiplied by 2 [1]. So the turning point $(6, -2)$ maps to $(12, -2)$ [1].
Q20 (3 marks): The coefficient 3 outside the function gives a vertical dilation by factor 3, and the negative sign gives a reflection in the $x$-axis [1]. $f(x + 1)$ is a translation 1 unit to the left [1]. The $-4$ is a translation 4 units down [1].
Q21 (4 marks): Start at $(4, 6)$. Translate right 3 (replace $x$ with $x - 3$, so add 3 to the $x$-coordinate): $(7, 6)$ [1]. Vertical dilation by factor 2 (multiply the $y$-coordinate by 2): $(7, 12)$ [1]. Translate up 1 (add 1 to the $y$-coordinate): $(7, 13)$ [1]. Therefore the image is $(7, 13)$ [1].
Q22 (4 marks): (a) Comparing with $y = a(x - h)^2 + k$ where $a = -2$, $h = 3$, $k = 8$: vertical dilation by factor 2, reflection in the $x$-axis, translation 3 units right, and translation 8 units up [1]. (b) The vertex is at $(3, 8)$; since $a = -2 < 0$ the parabola opens downward, so it is a maximum [1]. (c) $y$-intercept: substitute $x = 0$: $y = -2(0 - 3)^2 + 8 = -2(9) + 8 = -18 + 8 = -10$ [1], so the $y$-intercept is $(0, -10)$ [1].
Q23 (4 marks): Group and move the constant: $x^2 + 6x + y^2 - 8y = -9$ [1]. Complete both squares, adding 9 and 16 to each side: $(x + 3)^2 + (y - 4)^2 = -9 + 9 + 16 = 16$ [1]. Centre $(-3, 4)$ and radius 4 [1]. Domain $[-7, 1]$ and range $[0, 8]$ [1].