Quadratic Functions and Parabolas: Practice

Equal outputs at $2$ and $10$ make them mirror partners, so the axis lies at their midpoint. A positive leading coefficient opens the parabola upward, putting the minimum on the axis.

$x = \frac{2 + 10}{2} = 6$

The minimum occurs at $x = 6$.

For an upward parabola the height grows with distance from the axis, by the identity $f(6 + t) = f(6) + at^2$ with $a > 0$. Compare distances.

$|9 - 6| = 3 > 2 = |4 - 6|$

The input $9$ is farther from the axis, so $f(9) > f(4)$.

Substitute the known point $(0, 5)$ into the vertex form and solve for $a$.

$5 = a(0 - 3)^2 - 4 \ \Rightarrow\ 9a = 9 \ \Rightarrow\ a = 1$

So $a = 1$.

The axis is $x = -\dfrac{8}{2(-2)} = 2$, and since $a = -2 < 0$ the vertex value is a maximum.

$f(2) = -8 + 16 - 3 = 5$

The outputs run up to $5$, so the range is $(-\infty, 5]$.

The range $[3, \infty)$ means the smallest output is $3$, so every output is at least $3$.

$y \ge 3 > 0 \ \Rightarrow\ y = 0 \text{ never}$

The graph never touches the $x$-axis, so it has no real zeros.

The two intercepts are zeros, an equal-height pair at height $0$, so the axis bisects them.

$x = \frac{2 + 8}{2} = 5$

The axis of symmetry is $x = 5$.

The axis is $x = 4$, and the vertex height decides everything since the parabola opens upward.

$f(4) = 16 - 32 + 20 = 4 > 0$

The lowest point sits at height $4$, above the axis, so the graph never crosses it: no real zeros.

The vertex sits at $x = -\dfrac{b}{2a}$; putting it on the $y$-axis means that input is $0$.

$-\frac{b}{2a} = 0 \ \Rightarrow\ b = 0$

So $b = 0$; the value of $c$ only sets the vertex's height, not its horizontal position.

Factor out the highest power: $f(x) = x^2\left(a + \dfrac{b}{x} + \dfrac{c}{x^2}\right)$. For large $|x|$ the bracket approaches $a > 0$ while $x^2$ grows without bound.

$a > 0 \ \Rightarrow\ f(x) \to +\infty \text{ on both ends}$

Both ends of the parabola rise to $+\infty$.

The extremum sits on the axis $x = -\dfrac{b}{2a}$, with $a = -4$ and $b = 16$.

$x = -\frac{16}{2(-4)} = -\frac{16}{-8} = 2$

Since $a < 0$ this is a maximum, at $x = 2$.

The maximum is the vertex height, evaluated at the axis $x = 2$.

$f(2) = -4(4) + 16(2) - 7 = -16 + 32 - 7 = 9$

The maximum value is $9$.

The zeros $r_1$ and $r_2$ are a pair of inputs sharing the output $0$, so they are mirror partners and the axis bisects them.

$x = \frac{r_1 + r_2}{2}$

The axis lies at the average of the two roots.