Quadratic Functions and Parabolas: Additional Practice (Set 2)

The axis is $x = -\dfrac{b}{2a}$. Set it equal to $2$ and solve.

$-\frac{b}{2 \cdot 3} = 2 \ \Rightarrow\ -b = 12 \ \Rightarrow\ b = -12$

So $b = -12$.

The $y$-intercept is $f(0)$, and vertex form evaluates just as readily at $0$.

$f(0) = 3(0 - 2)^2 - 5 = 3(4) - 5 = 7$

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

Substitute $t = 3$ into the displacement identity.

$f(m + 3) = -5 + 2(3)^2 = -5 + 18 = 13$

The output $3$ units from the axis is $13$.

The axis is $x = -3$, and the vertex height settles it since the parabola opens upward.

$f(-3) = 9 - 18 + 11 = 2 > 0$

The lowest point sits at height $2$, above the $x$-axis, so there are no real zeros.

Standard form shows the $y$-intercept $c$ and, through the sign of $a$, the opening direction, all at sight. The vertex is hidden.

$\text{vertex } x = -\frac{b}{2a}, \quad y = f\!\left(-\frac{b}{2a}\right)$

Finding the vertex takes a division and an evaluation, so it is the feature that requires computation.

Squaring, scaling, and adding accept every real input, so the domain is always all reals. The other claims fail: the range is bounded on one side, zeros may be $0$, $1$, or $2$, and evenness needs $b = 0$.

$\text{domain of } ax^2 + bx + c = (-\infty, \infty)$

So the guaranteed property is that the domain is all real numbers.

For a downward parabola the vertex value $k$ is the maximum, so $k = 9$. Every output lies at or below it.

$-(x - 2)^2 \le 0 \ \Rightarrow\ y \le 9$

So $k = 9$ and the range is $(-\infty, 9]$.

The factored form shows zeros at $1$ and $5$, a mirror pair at height $0$, so the axis bisects them.

$x = \frac{1 + 5}{2} = 3$

The axis of symmetry is $x = 3$.

The two points share the height $5$, so they are mirror partners and the axis bisects them.

$x = \frac{1 + 9}{2} = 5$

The axis of symmetry is $x = 5$.

The opening direction is set by the coefficient of $x^2$: positive opens up, negative opens down.

$a > 0 \ \Rightarrow\ \text{up}, \qquad a < 0 \ \Rightarrow\ \text{down}$

So the sign of the leading coefficient $a$ decides it, while $c$, $k$, and the axis only place the graph.

Completing the square rewrites $ax^2 + bx + c$ as a perfect square plus a constant.

$ax^2 + bx + c \ \longrightarrow\ a(x - h)^2 + k$

That is exactly standard form turned into vertex form.

The vertex can sit above, on, or below the axis, giving three cases for an upward parabola (and symmetrically for a downward one).

$\text{meetings} \in \{0, 1, 2\}$

A parabola can miss the axis, touch it once at the vertex, or cross it twice, so $0$, $1$, or $2$.