Graphing from the Vertex Form

Student Summary

Not surprisingly, vertex form is especially helpful for finding the vertex of a graph of a quadratic function. For example, we can tell that the function, pp, given by p(x)=(x3)2+1p(x) = (x-3)^2 + 1 has a vertex at (3,1)(3,1).

We also noticed that, when the squared expression (x3)2(x-3)^2 has a positive coefficient, the graph opens upward. This means that the vertex, (3,1)(3,1), represents the minimum function value, p(x)p(x).

<p>Parabola in x y plane, origin O.</p>
Parabola in x y plane, origin O. X axis 0 to 6, by 1’s. Y axis 0 to 10, by 2s. Parabola opens upward with vertex at 3 comma 1. Left side of parabola crosses the y axis at 10.

But why does function pp take on its minimum value when xx is 3?

Here is one way to explain it: When x=3x=3, the squared term (x3)2(x-3)^2 equals 0, because (33)2=02=0(3-3)^2 =0^2=0. When xx is any other value besides 3, the squared term (x3)2(x-3)^2 is a positive number greater than 0. (Squaring any number results in a positive number.) This means that the output when x3x \neq 3 will always be greater than the output when x=3x=3, so function pp has a minimum value at x=3x=3.

This table shows some values of the function for some values of xx. Notice that the output is the least when x=3x=3, and it increases both as xx increases and as it decreases.

xx 0 1 2 3 4 5 6
(x3)2+1(x-3)^2+1 10 5 2 1 2 5 10

The squared term sometimes has a negative coefficient, for instance in h(x)=-2(x+4)2h(x)= \text-2(x+4)^2. The xx value that makes (x+4)2(x+4)^2 equal 0 is -4, because (-4+4)2=02=0(\text-4+4)^2=0^2=0. Any other xx value makes (x+4)2(x+4)^2 greater than 0. But when (x+4)2(x+4)^2 is multiplied by a negative number like -2, the resulting expression, -2(x+4)2\text-2(x+4)^2, ends up being negative. This means that the output when x -4x \neq  \text-4 will always be less than the output when x=-4x=\text-4, so function hh has its maximum value when x=-4x =\text-4.

<p>Parabola in x y plane, origin O. X axis negative 7 to 0, by 1’s. Y axis negative 8 to 2, by 2s. Parabola opens downward with vertex at negative 4 comma 0.</p>

Remember that we can find the yy-intercept of the graph representing any function that we have seen. The yy-coordinate of the yy-intercept is the value of the function when x=0x = 0. If gg is defined by g(x)=(x+1)25g(x)=(x+1)^2-5, then the yy-intercept is (0,-4)(0,\text-4) because g(0)=(0+1)25=-4g(0)=(0+1)^2 -5=\text-4. Its vertex is at (-1,-5)(\text-1,\text-5). Another point on the graph with the same yy-coordinate is located the same horizontal distance from the vertex but on the other side.

<p>Parabola in x y plane, origin O.</p>
Parabola in x y plane, origin O. X axis negative 4 to 1, by 1’s. Y axis negative 8 to 2, by 2s. Parabola opens upward with vertex at negative 1 comma negative 5. Other points on the parabola given are negative 2 comma negative 4 and 0 comma negative 4.

Visual / Anchor Chart

Standards

Addressing
F-IF.C

HSF-IF.C

F-IF.7.a

F-IF.7.a

F-IF.7.a

F-IF.7.a

HSF-IF.C.7.a

F-IF.7.a

F-IF.7.a

F-IF.7.a

F-IF.7.a

HSF-IF.C.7.a