Question

describe the transformation from the graph of f to the graph of g.

1. graph of f and g(x)=a·f(x) shown
2. table with x, f(x), g(x) values: x: -2, -1, 0, 1; f(x): 26, 2, -6, 2; g(x): 34, 10, 2, 10

write a rule for g.

1. the graph of g is a horizontal shrink by a factor of 2/3, followed by a translation 5 units left and 2 units down of the graph of f(x)=x².
2. the graph of g is a translation 2 units left and 3 units up, followed by a reflection in the y - axis of the graph of f(x)=x² - 2x.
3. the graph represents the path of a football kicked by a player, where x is the horizontal distance (in yards) and y is the height (in yards). the player kicks the ball a second time so that it travels the same horizontal distance, but reaches a maximum height that is 6 yards greater than the maximum height of the first kick. write a function that models the path of the second kick. graph with y=-0.02(x - 30)²+18 shown

Explanation:

Step 1: Analyze problem 7

For a horizontal shrink of $y = f(x)$ by a factor of $\frac{2}{3}$, we replace $x$ with $\frac{3}{2}x$ (since if $y = f(x)$ and we want to shrink horizontally by a factor $k$, the new function is $y=f(kx)$). Then for a translation 5 units left, we replace $x$ with $(x + 5)$ and for 2 units down we subtract 2 from the function. Given $f(x)=x^{2}$, the new function $g(x)$ is $g(x)=(\frac{3}{2}(x + 5))^{2}-2=\frac{9}{4}(x + 5)^{2}-2$.

Step 2: Analyze problem 8

First, for the translation of $y = f(x)=x^{2}-2x$ 2 units left and 3 units up. For a left - translation of 2 units, we replace $x$ with $(x + 2)$ and for an up - translation of 3 units we add 3 to the function. So we get $y=(x + 2)^{2}-2(x + 2)+3$. Expand it: $y=x^{2}+4x + 4-2x-4 + 3=x^{2}+2x + 3$. Then for a reflection in the $y$ - axis, we replace $x$ with $-x$. So $g(x)=(-x)^{2}+2(-x)+3=x^{2}-2x + 3$.

Step 3: Analyze problem 9

The function of the first - kick is $y=-0.02(x - 30)^{2}+18$. The second kick travels the same horizontal distance but has a maximum height 6 yards greater. The vertex form of a parabola is $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex. Here $a=-0.02$ and $h = 30$ remains the same, and $k$ increases by 6. So the function for the second kick is $y=-0.02(x - 30)^{2}+18 + 6=-0.02(x - 30)^{2}+24$.

Answer:

1. $g(x)=\frac{9}{4}(x + 5)^{2}-2$
2. $g(x)=x^{2}-2x + 3$
3. $y=-0.02(x - 30)^{2}+24$