Scribd
Upload
78 views4 pages

Quadratic Functions: Graphing & Inverses

This document discusses quadratic functions and their graphs. It provides 3 examples: 1) Graphing a quadratic function in the form f(x) = -2(x+1)2 + 3 and finding its vertex. The vertex is (-1,3). 2) Finding the intercepts of f(x) = -2(x+1)2 + 3. The y-intercept is 1. The x-intercepts are approximately -1.2 and 0.2. 3) Finding the inverse of the quadratic function f(x) = 2(x - 2)2 + 3 for x ≤ 2. The inverse function is f^-1(x) = 2 - √[(

Uploaded by

Mathias Mike
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
78 views4 pages

Quadratic Functions: Graphing & Inverses

This document discusses quadratic functions and their graphs. It provides 3 examples: 1) Graphing a quadratic function in the form f(x) = -2(x+1)2 + 3 and finding its vertex. The vertex is (-1,3). 2) Finding the intercepts of f(x) = -2(x+1)2 + 3. The y-intercept is 1. The x-intercepts are approximately -1.2 and 0.2. 3) Finding the inverse of the quadratic function f(x) = 2(x - 2)2 + 3 for x ≤ 2. The inverse function is f^-1(x) = 2 - √[(

Uploaded by

Mathias Mike
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 4

SUMMARY

Lines, Parabolas, and Systems

3.3 Quadratic Functions

Example 1: Graphing a Quadratic Function


DEFINITION:
A quadratic function is a function of the form
Describe the graph of the quadratic function
f (x) = ax2
+ bx + c
with a  0 , so its formula involves a square term as well as linear and constant terms. We
call this form the standard form F(x)
of the = formula for a2 quadratic
-2(x+1) +3 function.

SOLUTION:

Since a = -2 < 0, the graph is a parabola that opens downward. Because |-


2| > 1, the graph is a narrower than the graph of the squaring function. In
order to determine the vertex, this function rule needs to be written in the
vertex form f(x) = a(x-h)2 + k. Since we can replace adding 1 with
subtracting -1, the proper vertex form for this function is f(x) = -2x(x-(-
1))2 + 3. Thus the vertex of the parabola is the point (-1,3)

The Quadratic Formula


Figure 3.1 : The graphs of the
squaring function and f(x) =
Suppose that our quadratic function f(x) is-2(x+1)
given 2 in standard form. That
+3
is,

f(x) = ax2 + bx + c

Then solving the equation f(x) = 0 means solving the quadratic equation

Ax2 +bx + c = 0
This equation has two solutions r and s given by
 b  b 2  4 ac  b  b 2  4ac
r= 2a and s = 2a

This fact that both values x = r and x = s given by these two formulas
satisfy the given quadratic equation is what is meant by the quadratic
formula written in usual form which describes both solutions
simultaneously

 b  b 2  4ac
X= 2a

Example 2 : Finding the intercepts of the graph

Find the intercepts of the graph of the quadratic function

F(x) = -2(x+1)2 + 3

Solution

Since any real value of x will give a sensible value for f(x), the domain of
f is all real numbers. So f(0) exists, find the y-intercept by evaluating

F(0) = -2(0+1)2 + 3 = -2 + 3 = 1

The y-intercept, then, is 1.

To find any x-intercepts, solve the equation -2(x+1) + 3 = 0. In order to


do so with the quadratic formula, the quadratic expression must be in
standard form, not vertex form. So, expanding the squared binomial and
simplifying will get:

-2(x2 + 2x +1) + 3 = 0,
-2x2- 4x - 2 + 3 = 0,
-2x2 - 4x + 1 = 0

Then with a = -2, b = -4, and c = 1, the quadratic formula gives

 (4)  (4) 2  4(2)(1)


X= 2( 2)
4  16  8
= 4

4 2 6
= 4

4  24
= 4

So the two solution of quadratic equation f(x) = 0 are

4 2 6 6 42 6 6
 1   2.2  1   0.2
4 2 and  4 2

Example 3

Find the inverse of the quadratic function in vertex form given by

f(x) = 2(x - 2) 2 + 3 , for x <= 2

Write the function as an equation :

y = 2(x - 2) 2 + 3

Solve the above for x to obtain 2 solutions

(x - 2) 2 = (y - 3) / 2

x - 2 = + or - √[ (y - 3)/2 ]

x = 2 + √[ (y - 3)/2 ]

and

x = 2 - √[ (y - 3)/2 ]

Since x given by x = 2 - √[ (y - 3)/2 ] is always less than or equal to 2, we


take the solution.

x = 2 - √[ (y - 3)/2 ]

Change x into y and y into x to obtain the inverse function.

y = 2 - √[ (x - 3)/2 ]

f -1(x) = 2 - √[ (x - 3)/2 ]

You might also like