Analyzing Graphs of Quadratic Functions

There are two important topics in this section: graphing the quadratic function $f(x)=ax^2+bx+c$ and finding the (absolute) maximum or the minimum value of $f(x)=ax^2+bx+c$.

First the sign of the leading coefficient $a$ tells us some information about the graph. If $a>0$ then the tail of the graph goes up, i.e. the graph is a smiling face $\smile$. If $a<0$ then the tail of the graph goes down, i.e. the graph is a frowning face $\frown$.

Using the completing the square $f(x)=ax^2+bx+c$ can be written as
$$f(x)=a(x-h)^2+k,$$
where $h=-\frac{b}{2a}$ and $k=f(h)=f\left(-\frac{b}{2a}\right)$. The ordered pair $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ is called the vertex of the parabola $f(x)$ and the vertical line $x=-\frac{b}{2a}$ is called the axis of symmetry (this is the vertical line that divides the graph of $f(x)$ into two halves). If $a>0$, then $f\left(-\frac{b}{2a}\right)$ is the absolute minimum value of $f(x)$. If $a<0$, then $f\left(-\frac{b}{2a}\right)$ is the absolute maximum value of $f(x)$.

How to sketch the graph of $f(x)=a(x-h)^2+k$?

Your textbook is telling you to sketch the graph of $f(x)=a(x-h)^2+k$ using transformations that you learned in section 1.7 (here and here). In principle, it is right to use transformations but in practice there is an easier way to do. All you need is the sign of $a$, the vertex $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$, and the $y$-intercept $c$. (Although not required, it would be better if you know $x$-intercepts as well.)

Example. Let $f(x)=x^2+7x-8$.

(a) Find the vertex.

Solution. $-\frac{b}{2a}=-\frac{7}{2}$ and
\begin{align*}
f\left(-\frac{b}{2a}\right)&=f\left(-\frac{7}{2}\right)\\
&=\left(-\frac{7}{2}\right)^2+7\left(-\frac{7}{2}\right)-8\\
&=-\frac{81}{4}.
\end{align*}

(b) Find the axis of symmetry.

Solution. The axis of symmetry is the vertical line $x=-\frac{b}{2a}=-\frac{7}{2}$.

(c) Determine whether there is a maximum or minimum value and find that value.

Solution. Since $a=1>0$, there is a minimum and the minimum value is the $y$-coordinate of the vertex $f\left(-\frac{7}{2}\right)=-\frac{81}{4}$.

(d) Graph the function.

Solution. Since $a=1>0$, the graph is a parabola that opens up (smiling face). Also note that the $y$-intercept of $f(x)$ is $-8$. In fact, we can extract more information since $f(x)$ can be easily factored as $(x+8)(x-1)$, so the $x$-intercepts are $x=-8,1$.

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