Stretching and Reflecting Graphs: Additional Practice (Set 2)

The inside $2x$ reaches the old input at half the $x$-value, and the $-3$ lowers the output.

$\left(\tfrac{p}{2},\; q - 3\right)$

Set the output to $0$ and solve for $x$.

$3(x + 2)^2 = 12 \;\Rightarrow\; (x + 2)^2 = 4 \;\Rightarrow\; x = 0 \text{ or } x = -4$

The crossings are $(0, 0)$ and $(-4, 0)$.

At $x = 3$ the stretched graph has height $2f(3)$, and that equals $10$.

$2f(3) = 10 \;\Rightarrow\; f(3) = 5$

At $x = 6$ the compressed graph has height $\tfrac{1}{3}f(6)$, and that equals $4$.

$\tfrac{1}{3}f(6) = 4 \;\Rightarrow\; f(6) = 12$

With vertex at the origin the equation is $y = ax^2$. Use the point $(1, -3)$ to find $a$.

$-3 = a(1)^2 \;\Rightarrow\; a = -3$

The negative value makes it open downward, so $y = -3x^2$.

If $(a, b)$ is on $y = f(2x)$ then $b = f(2a)$, so the input $2a$ produces height $b$ on the original.

$(2a,\; b) \text{ lies on } y = f(x)$

Reflecting across the y-axis negates the input, then the vertical stretch doubles the height.

$(4, -1) \longrightarrow (-4, -1) \longrightarrow (-4, -2)$

A horizontal stretch by $3$ scales each input by $3$, which needs $b = \tfrac{1}{3}$ inside the function.

$y = f\left(\tfrac{x}{3}\right)$

Multiplying the input by $3$ would compress it, and factors outside change the height.

Negating every output flips the range across $0$, so each endpoint changes sign and the ends swap order.

$-4 \le y \le 2 \;\longrightarrow\; -2 \le y \le 4$

The inside is $-\tfrac{x}{2}$, so $b = -\tfrac{1}{2}$. The negative sign reflects across the y-axis, and the size $\tfrac{1}{2}$ stretches inputs by $\tfrac{1}{|b|} = 2$.

$b = -\tfrac{1}{2} \;\Rightarrow\; \text{reflect across y-axis, stretch by } 2$

The intercept at $x = 3$ means the inside $2x$ hits a zero of $f$ there.

$2 \times 3 = 6 \;\Rightarrow\; f(6) = 0$

So the original graph has its x-intercept at $(6, 0)$.

Reflecting across the y-axis negates each input and keeps each height.

$(3, 5) \to (-3, 5), \qquad (-1, 5) \to (1, 5)$