Lesson 13: Graphing the Standard Form (Part 2)

Graphing the Standard Form (Part 2)

Student Summary

In an earlier lesson, we saw that a quadratic function written in standard form, $ax^2 + bx +c$ , can tell us some things about the graph that represents it. The coefficient $a$ can tell us whether the graph of the function opens upward or downward, and also gives us information about whether it is narrow or wide. The constant term $c$ can tell us about its vertical position.

Recall that the graph representing $y = x^2$ is an upward-opening parabola with the vertex at $(0,0)$ . The vertex is also the $x$ -intercept and the $y$ -intercept.

Suppose we add 6 to the squared term: $y=x^2+6$ . Adding a 6 shifts the graph upward, so the vertex is at $(0,6)$ . The vertex is the $y$ -intercept, and the graph is centered on the $y$ -axis.

<p>Coordinate plane, 2 graphs of quadratic functions. First, y = x squared has minimum at the origin. Next, y = x squared + 6 has minimum at 0 comma 6.</p>

What can the linear term $bx$ tell us about the graph representing a quadratic function?

The linear term has a somewhat mysterious effect on the graph of a quadratic function. The graph seems to shift both horizontally and vertically. When we add $bx$ (where $b$ is not 0) to $x^2$ , the graph of $y=x^2+bx$ is no longer centered on the $y$ -axis.

Suppose we add $6x$ to the squared term: $y=x^2+6x$ . Writing the $x^2+6x$ in factored form as $x(x+6)$ gives us the zeros of the function, 0 and -6. Adding the term $6x$ seems to shift the graph to the left and down and the $x$ -intercepts are now $(\text-6,0)$ and $(0,0)$ . The vertex is no longer the $y$ -intercept, and the graph is no longer centered on the $y$ -axis.

<p>Coordinate plane, 2 graphs of quadratic functions. First, y = x squared has minimum at the origin. Next, y = x squared + 6 x has minimum at negative 3 comma negative 9.</p>

What if we add $\text-6x$ to $x^2$ ? We know that $x^2-6x$ can be rewritten as $x(x-6)$ , which tells us the zeros: 0 and 6. Adding a negative linear term to a squared term seems to shift the graph to the right and down. The $x$ -intercepts are now $(0,0)$ and $(6,0)$ . The vertex is no longer the $y$ -intercept, and the graph is not centered on the $y$ -axis.