Parabolas: Practice

Use $x = -\tfrac{b}{2a}$ with $a = 2$ and $b = 8$.

$x = -\frac{8}{2(2)} = -\frac{8}{4} = -2$

So the axis of symmetry is $x = -2$.

The vertex x-coordinate is $-\tfrac{b}{2a}$ with $a = 1$, $b = -6$.

$x = -\frac{-6}{2} = 3$

Substitute back for the height: $y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$. So the vertex is $(3, -4)$.

Complete the square: half of $4$ is $2$, and $2^2 = 4$, so add and subtract $4$.

$y = (x^2 + 4x + 4) - 4 + 1 = (x + 2)^2 - 3$

The perfect square is $(x + 2)^2$ and the leftover constant is $-4 + 1 = -3$.

Set $y = 0$ and factor, looking for two numbers that multiply to $-6$ and add to $-1$: those are $-3$ and $2$.

$x^2 - x - 6 = (x - 3)(x + 2) = 0$

So $x = 3$ or $x = -2$, giving the x-intercepts at $x = 3, -2$.

Set $y = 0$ and factor: two numbers that multiply to $6$ and add to $-5$ are $-2$ and $-3$.

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

So $x = 2$ or $x = 3$.

Points that are mirror images across the axis of symmetry share the same height. The point $x = 1$ is $3$ units left of the axis $x = 4$, so its mirror is $3$ units to the right, at the same height $y = 7$.

$4 + (4 - 1) = 7 \;\Rightarrow\; (7, 7)$

So $(7, 7)$ must also lie on the parabola.

Use $x = -\tfrac{b}{2a}$ with $a = -1$ and $b = 4$, watching the negative $a$.

$x = -\frac{4}{2(-1)} = -\frac{4}{-2} = 2$

So the axis of symmetry is $x = 2$.

Set $y = 0$. This is a difference of squares, $x^2 - 9 = (x - 3)(x + 3)$.

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

So the x-intercepts are $x = 3$ and $x = -3$.

Half of $10$ is $5$, and $5^2 = 25$, so add and subtract $25$.

$y = (x^2 + 10x + 25) - 25 = (x + 5)^2 - 25$

The square is $(x + 5)^2$ and the leftover is $-25$, so the vertex is $(-5, -25)$.

Find $x = -\tfrac{b}{2a}$ with $a = 3$, $b = -12$.

$x = -\frac{-12}{2(3)} = \frac{12}{6} = 2$

Substitute: $y = 3(2)^2 - 12(2) + 7 = 12 - 24 + 7 = -5$. The vertex is $(2, -5)$.

A downward parabola has its vertex as the highest point, so the maximum value of $y$ is the vertex's y-coordinate.

$\text{vertex } (2, 9) \;\Rightarrow\; \text{maximum } y = 9$

Every other point sits below height $9$.

Set $y = 0$ and factor out the common $x$.

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

So the x-intercepts are $x = 0$ and $x = -3$. A missing constant term means the origin is one of them.