Quadratic Functions and Parabolas: Practice

A quadratic function has the form $ax^2 + bx + c$ with $a \ne 0$: the highest power of $x$ is exactly $2$.

$f(x) = 2x^2 - 5x + 1$

The first option is linear (highest power $1$), the third is not a polynomial, and the fourth has a cube, so only the second is quadratic.

The leading coefficient is the number multiplying $x^2$ in $ax^2 + bx + c$.

$a = 4, \quad b = -7, \quad c = 2$

So $a = 4$; the $-7$ is $b$ and the $2$ is $c$.

The constant term is the number with no $x$ attached, which is also the output $f(0)$.

$c = f(0) = -9$

The $5$ is $b$ and the coefficient of $x^2$ is $-1$, so $c = -9$.

The whole bend of a parabola lives in the $x^2$ term. Setting $a = 0$ removes it.

$0 \cdot x^2 + bx + c = bx + c$

That is a straight line, not a parabola, so $a \ne 0$ is what keeps the function quadratic.

Vertex form $a(x - h)^2 + k$ has vertex $(h, k)$. Match the parts, remembering the inside runs backward.

$x - 4 = x - h \ \Rightarrow\ h = 4, \qquad k = 3$

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

The axis of symmetry of $a(x - h)^2 + k$ is the vertical line through the vertex, $x = h$.

$h = 4 \ \Rightarrow\ x = 4$

The axis is the vertical line $x = 4$, passing through the vertex $(4, 3)$.

Write the inside as a subtraction to read $h$ correctly.

$x + 2 = x - (-2) \ \Rightarrow\ h = -2, \qquad k = -5$

So the vertex is $(-2, -5)$: the plus sign inside gives a negative $h$.

The direction is set by the sign of the leading coefficient $a$.

$a = -3 < 0 \ \Rightarrow\ \text{opens downward}$

A parabola of this form always opens up or down, never sideways, and a negative $a$ opens it down.

Here $a = -1 < 0$, so the parabola opens downward and its vertex is the highest point.

$-(x-5)^2 \le 0 \ \Rightarrow\ y \le 8$

The maximum value is $k = 8$, reached at $x = 5$.

Substituting $x = 0$ kills every term that carries an $x$.

$f(0) = 2(0) + 3(0) - 1 = -1$

So $f(0) = -1$, which is just the constant term $c$.

The parabola opens upward ($a = 1 > 0$) with vertex value $k = 4$, so $4$ is the smallest output.

$(x - 2)^2 \ge 0 \ \Rightarrow\ y \ge 4$

The outputs fill $[4, \infty)$.

Read off $a = 1$ and $b = -4$, then apply the formula, watching the signs.

$x = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2$

The axis of symmetry is $x = 2$.