Quadratic Functions and Parabolas: Practice

Here $a = 1$ and $b = -6$, so use $x = -\dfrac{b}{2a}$.

$x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3$

The axis of symmetry is $x = 3$.

The axis is $x = -\dfrac{-6}{2} = 3$, and one evaluation gives the height.

$f(3) = 9 - 18 + 5 = -4$

So the vertex is $(3, -4)$.

With $a = 2 > 0$ the parabola opens upward, so its vertex value is the minimum. The axis is $x = -\dfrac{-8}{4} = 2$.

$f(2) = 8 - 16 + 1 = -7$

The minimum output is $-7$, so the range is $[-7, \infty)$.

The $y$-intercept is $f(0)$, the constant term in standard form.

$f(0) = 0 - 0 + 7 = 7$

So the graph meets the $y$-axis at $(0, 7)$.

Factored form $a(x - r_1)(x - r_2)$ displays its zeros directly: the value in each factor is where that factor vanishes, so read $r_1 = 3$ from $(x - 3)$ and $r_2 = -5$ from $(x + 5)$. Confirm by substituting each back in.

$f(3) = (0)(8) = 0, \qquad f(-5) = (-8)(0) = 0$

Both inputs give output $0$, so the zeros are $3$ and $-5$.

The zeros are $1$ and $7$, an equal-height pair (both at height $0$), so the axis bisects them.

$x = \frac{1 + 7}{2} = 4$

The axis of symmetry is $x = 4$.

Each form answers one question for free. The vertex $(h, k)$ is read straight off the completed square.

$a(x - h)^2 + k \ \Rightarrow\ \text{vertex } (h, k)$

Standard form hides the vertex and factored form shows the zeros, so vertex form is the one that displays it.

Since $a = -1 < 0$ the parabola opens downward, so the vertex value is the maximum. The axis is $x = 2$.

$f(2) = -4 + 8 - 1 = 3$

The maximum value is $3$.

Two inputs with the same output are mirror partners, so the axis stands halfway between them.

$x = \frac{1 + 7}{2} = 4$

The axis of symmetry is $x = 4$, with no coefficients needed.

The parabola opens upward ($a = 3 > 0$), so the vertex value $k$ is the minimum.

$3(x + 1)^2 \ge 0 \ \Rightarrow\ y \ge -12$

The minimum value is $-12$, attained at $x = -1$.

Opening upward from vertex $(0, 4)$, the lowest point sits at height $4$, above the $x$-axis.

$y \ge 4 > 0 \ \Rightarrow\ y \ne 0 \text{ ever}$

The graph never reaches height $0$, so it has no $x$-intercepts.

Here $a = 2$ and $b = 12$, so the axis is $x = -\dfrac{12}{4} = -3$.

$f(-3) = 2(9) + 12(-3) + 19 = 18 - 36 + 19 = 1$

So the vertex is $(-3, 1)$.