Algebra 1, Please Help! Graphing from Factored Form

Question

1 A function f is given by f(x)=(x−4)(x+2)

Answer the following and explain your technique/reasoning:

1. What are the x-intercepts of the graph?

2. What is the ordered pair of the vertex of the graph?

3. What is the y-intercept of the graph?

2 Based on your answers to #1, make the graph of the function.

Answer 1

f(x) = (x - 4)(x +2)

1. x-intercepts occur at f(x) or y = 0. Then either x - 4 = 0 or x + 2 = 0, or both.

So x = 4 or x = -2

2. The x-value of the vertex of a parabola is -b/2a when function has form ax² + bx + c.

We need to get the function into that form (use FOIL): x² - 2x - 8. So x-value is -(-2)/2(1) = 1.

Can substitute x = 1 into the equation to find corresponding y value: 1² - 2(1) - 8 = -9.

Vertex is at (1, -9)

3. The y-int occurs where x = 0. Can use original form: (0 - 4)(0 + 2) = -8, at (0, -8)

You have enough info to draw the parabola. It opens upwards because "a" is positive. You have both x-intercepts, the ordered pair for the vertex and the y-intercept.

Answer 2

To find the x-intercepts, you make "y" equal to zero:

0 = (x − 4)(x + 2)

Here, you are multiplying two things ("x - 4" and "x + 2" and you are getting zero as the result. The only way that can happen is if one of the two things IS zero. So, you can set each "thing" equal to zero and solve. First x-intercept: x - 4 = 0 or x = 4. Second x-intercept: x + 2 = 0 or x = -2. Please note that for x-intercepts, the y-value is zero so the points are (4, 0) and (-2, 0).

If you have the x-intercepts, you can find the x-value of the vertex by averaging them: (4 + -2)/2 = 1. To find the y-value of the vertex, plug in x = 1 into the function: f(1) = (1 - 4)(1 + 2) = -9 so the vertex is (1, -9)

To find the y-intercept, make x equal to zero:

f(0) = (0 - 4)(0 + 2) = -8

Note that the x-value is zero so the point is (0, -8)Then the graph looks like this: