7.11.3 Sketching a Graph of a Quadratic Function Using at Least Three Identifiable Points

Graphing Quadratics With Points

The function $f$ given by $f(x)=(x+1)(x-1)$ is written in factored form. Recall that this form is helpful for finding the zeros of the function (where the function has the value 0) and telling us the $x$-intercepts on the graph representing the function.

Here is a graph representing $f$. It shows two $x$-intercepts at $x=-1$ and $x=3$.

If we use –1 and 3 as outputs to $f$, what are the outputs?

• $f(-1)=(-1+1)(-1-3)=(0)(-4)=0$
• $f(3)=(3+1)(3-3)=(4)(0)=0$

Because the inputs –1 and 3 produce an output of 0, they are the zeros of the function $f$. And because both $x$ values have 0 for their $y$ value, they also give us the $x$-intercepts of the graph (the points where the graph crosses the $x$-axis, which always have a $y$-coordinate of 0). So, the zeros of a function have the same values as the $x$-coordinates of the $x$-intercepts of the graph of the function.

The factored form can also help us identify the vertex of the graph, which is the point where the function reaches its minimum value. Notice that the $x$-coordinate of the vertex is 1, and that 1 is halfway between –1 and 3. Once we know the $x$-coordinate of the vertex, we can find the $y$-coordinate by evaluating the function $f(1)=(1+1)(1-3)=(2)(-2)=-4$. So the vertex is at $(1,–4)$.

When a quadratic function is in standard form, the $y$-intercept is clear: its $y$-coordinate is the constant term $c$ in $a{x}^2+bx+c$. To find the $y$-intercept from factored form, we can evaluate the function at $x=0$, because the $y$-intercept is the point where the graph has an input value of 0:

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

Try it

Try It: Graphing Quadratics with Points

Find the zeros and vertex of the function $f$ given by $f(x)=(x-4)(x+2)$. Use the points to graph $f$.

Here is how to find the function’s zeros and vertex:

Step 1 - Set each factor equal to 0 and solve.

• $x-4=0,x=4$
• $x+2=0,x=-2$

Step 2 - Write the zeros as points. $(4,0)(-2,0)$

Step 3 - Find the $x$-value halfway between the zeros. $x=1$

Step 4 - Substitute $x=-1$ into the function to find the vertex. $f(1)=(1-4)(1+2)=(-3)(3)=-9$ Vertex: $(1,-9)$

Step 5 - Now, use these points to graph $f$: