Q. 102 - Precalculus Enhanced with Graphing Utilities

Question

Why does the graph of a quadratic function open up if $a>0$ and down if $a<0$?

Step-by-Step Solution

Answer

When $a>0$ then $f\left(x\right)$approaches positive infinity and hence it opens upwards

when $a<0$then $f\left(x\right)$ approaches negative infinity and hence it opens downwards. 

Step 1: Given information

Consider a quadratic equation $f\left(x\right)=a{(x-h)}^2+k$

Step 2: When a is positive

Now we evaluate,

$$\begin{gathered} f(\infty )=a{(\infty -h)}^2+k \\ f(\infty )=a{(\infty )}^2+k \\ f(\infty )=\infty \\ Also \\ f(-\infty )=a{(-\infty -h)}^2+k \\ f(-\infty )=a{(-\infty )}^2+k \\ f(-\infty )=\infty \end{gathered}$$

So when $a>0$ then $f\left(x\right)$approaches positive infinity

Step 3: When a is negative

Now we evaluate

$$\begin{gathered} f(\infty )=-a{(\infty -h)}^2+k \\ f(\infty )=-a{(\infty )}^2+k \\ f(\infty )=-\infty \\ Also \\ f(-\infty )=-a{(-\infty -h)}^2+k \\ f(-\infty )=-a{(-\infty )}^2+k \\ f(-\infty )=-\infty \end{gathered}$$

When $a<0$then $f\left(x\right)$approaches negative infinity

Step 4: Conclusion

When $a>0$ then $f\left(x\right)$ approaches positive infinity and hence it opens upwards Similarly when $a<0$ then $f\left(x\right)$approaches negative infinity and hence it opens downwards.