SOLUTION: 1. Given the following quadratic equation, find a. the vertex b. the axis of symmetry c. the intercepts d. the domain e. the range f. the interval where the function is

Question 784564: 1. Given the following quadratic equation, find
a. the vertex
b. the axis of symmetry
c. the intercepts
d. the domain
e. the range
f. the interval where the function is increasing, and
g. the interval where the function is decreasing
h. Graph the function. y = -x^2 + 2x + 8

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Terminology problem: The equation specified in step h below is NOT a quadratic equation. It is a quadratic function.

would be a quadratic equation. Quadratic equations have solutions over the complex numbers, but they do not have graphs, vertices, axes of symmetry, intercepts, domains, or ranges. Quadratic functions have all these things.

Given the function:

For your particular given function:

, and

The

-coordinate of the vertex is given by

and the

-coordinate of the vertex is given by

. So the vertex is

.

The axis of symmetry is

The

-coordinate of the

-intercept is found by substituting zero for

, hence the

-coordinate of the

-intercept is

, and therefore the

-intercept is the point

.

The

-coordinates of the

-intercepts, if they exist, are the zeros of the function. First, determine existence. If

, then the intercepts exist. Otherwise not. Hint: If

and

have opposite signs, then there are always two real and unequal zeros.

If the intercepts exist, then the

-coordinates of the

-intercepts are found by:

And the

-intercepts are the points

and

. In the case of a perfect square trinomial that has one zero with a multiplicity of 2, there is only one intercept point, and that point is identical to the vertex, see part a.

The domain of all polynomial functions is the set of real numbers, that is: