Question

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

Short 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-by-step solution

Given information

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

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

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

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.