ACCELERATED PRE-CALCULUS SUMMER STUDY GUIDE

CHAPTER 23: QUADRATICS AND EQUATIONS IN THE FORM OF y = ax² + bx + c

SECTION 23.1 WHAT IS A QUADRATIC SET EQUAL TO Y?

Equations written in the form of y = ax² + bx + c, where a, b and c are real numbers and a ≠ 0 are called "quadratics set equal to y." This type of "quadratic equation" will create ordered pairs that produce a figure called a "parabola."

AN ASIDE: YOU HAVE BEEN FACTORING QUADRATIC EXPRESSIONS ON MANY OCCASIONS!

Here are some examples of the above ideas:

ex. 1) Name a, b, and c, the vertex, and 4 ordered pairs of the parabola y = x² + 1x + 1 and then graph the ordered pairs and the vertex.

Let's get the vertex:

a is the coefficient of x² therefore a=1.

b is the coefficient of x: therefore b=1.

c is the numerical part, i.e. the isolated number with no variable part: therefore c = 1..

,

Let's get some ordered pairs and then create a graph:

ex. 2) Find a,b, and c; the vertex and 4 ordered pairs of the parabola y = -2x² + 1; graph the ordered pairs.

I can rewrite y = -2x² + 1 as y = -2x² + 0x + 1 so I can easily determine a, b, and c.

Let's get the vertex:

a is the coefficient of x² : therefore a = -2.

b is the coefficient of x: therefore b=0 since we don't have a term with x.

c is the numerical part, i.e. the isolated number with no variable part: therefore c = 1.

Here are the ordered pairs and the graph of the parabola:

SECTION 23.2 QUADRATICS AND X AND Y INTERCEPTS

In an earlier chapter, chapter 14 , I reintroduced the concept of x and y intercepts. I will look at the general rules for determining the x and y intercepts of a parabola.

ex.) Given y = x² + x - 6

a) Find the x and y intercept
b) Find the vertex
c) Plot the vertex and the x and y intercepts.

ANSWERS:
a) To find the x intercept set y = 0.

x + 3 = 0 or x - 2 = 0

-3 and 2 are the x intercepts.

The associated y values are 0.

The associated points are (-3, 0) and (2, 0)

To find the y intercept set x = 0.

-6 is the y intercept and its associated x value is 0.

The associated point is (0, -6)

b) y= x + x - 6 = 1x² + 1x + -6

a = 1: b = 1 ; c = -6

c)

Also note that since a=1 is positive the graph is concave up.

I will consider an example of a parabolic equation that produces a parabola with no x intercepts:

ex. Determine the x-intercepts for y = x² - 4x + 9

answer:

y = 1x² + -4x + 9

I rewrote the original quadratic as addition so I could get a, b, and c.

Therefore a = 1; b = -4; c = 9

To find the x - intercepts, set y = 0, then solve for x: 0 = x² - 4x + 9

We will use the Quadratic Formula to solve this since we can't factor  x² - 4x + 9.

Notice that the radicand is -20 and you can't find the square root of -20. What this reveals is:

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IF YOU GET A RADICAND THAT IS NEGATIVE THEN
YOU CAN CONCLUDE THERE ARE NO X-INTERCEPTS
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ex.) Graph y = x² - 4x + 9 by getting the vertex and the y intercept and sketch the graph to see if there are no x-intercepts.

Also note that since a = 1 is positive the graph is concave up.

a = 1; b = -4 ; c = 9

The vertex is: (2, 5)

To get the y-intercept: set x = 0 and solve for y:

Here's the graph:

example) Graph y = -3x² - 14x - 8 using the vertex and the x and y intercepts.

Note that since a = -3 is negative the graph will be concave down.

Note that I rewrote the above equation in terms of addition which is the form of the quadratic equation.

The vertex is:

y-intercept:

x-intercept:

or x + 4 = 0

or x = -4

HERE'S THE GRAPH: