Chapter 3.3, Problem 102AYU

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

To determine

The graph of a quadratic function opens up if $a>0$ and opens down if $a<0$.

Explanation of Solution

Given:

Consider the quadratic function $f(x)=a{(x-h)}^2+k$ with $a\ne 0$ and $(h,k)$ is the vertex of that function.

We have the function $f(x)=a{(x-h)}^2+k$.

Here, we can see that the shape of the graph is determined by the value $a{(x-h)}^2$.

The graph of $f(x)$ is the graph of the function $g(x)=a{x}^2$ with $k$ units shifted vertically and $h$ units shifted horizontally.

Therefore, in order to explain the graph of $f(x)$ it is enough to explain the graph of $g(x)$.

Thus, let us consider the function $g(x)=a{x}^2$.

We know that $x^2$ is always positive.

Therefore, the value of $g(x)=a{x}^2$ is always positive when $a$ is positive.

Therefore, the parabola should open upwards if it has to be positive always (so that it extends up to positive infinity).

Similarly, when $a$ is negative, we can see that the value of $g(x)=a{x}^2$ is negative.

Thus, the parabola has to open downward if it has to be negative always (as it extends up to negative infinity).

That is why the graph of a quadratic function opens up if $a>0$ and opens down if $a<0$.