SOLUTION: How do you compute the intercepts of a quadratic function?

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Question 39513: How do you compute the intercepts of a quadratic function?

Found 2 solutions by josmiceli, AnlytcPhil:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
y(x) = Ax^2 + Bx + C
y(0) will be the y-intercept. There should be just one.
This is found by setting x = 0 in the equation, so
y(0) = A(0) + B(0) + C
y(0) = C
The x-intercept(s) is found by setting y(x) = 0
There may be 0, 1, or 2 x-intercepts
Ax^2 + Bx + C = 0 solve for x
The formula is
x+=+%28-B+%2B-+sqrt%28+B%5E2-4%2AA%2AC+%29%29%2F%282%2AA%29+

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
How do you compute the intercepts of a quadratic function? 

Intercepts are points where a graph crosses the x axis or the y axis.

Every point on the x-axis has 0 for its y-coordinate.  And vice-versa, that is,

every point on the y-axis has 0 for its x-coordinate.

A quadratic graph  

1. Is either shaped like a U or an upside down U.

 +graph%28+88%2C+88%2C+-2%2C+6%2C+-4%2C+4%2C+x%5E2-4x%2B3%29+ or  +graph%28+88%2C+88%2C+-5%2C+3%2C+-4%2C+4%2C+-x%5E2-x%2B2%29+

2. Always has exactly 1 y-intercept

3. Sometimes has 2 x-intercepts, like the two graphed above

4.  Sometimes has NO x-intercepts, like this one just below:

 +graph%28+88%2C+88%2C+-2%2C+6%2C+-4%2C+4%2C+x%5E2-x%2B1%29+

5. Sometimes has exactly ONE x-intercept, like this one below

 +graph%28+88%2C+88%2C+-2%2C+6%2C+-4%2C+4%2C+-%28x-4%29%5E2%29+
 
6. Has as its general equation

     y = Ax² + Bx + C

     where A, B and C can represent positive or negative numbers,
     
7.  B and/or C is sometimes 0, but A is never 0. 


How to find the y-intercept:

Substitute 0 for x and solve for y:

Then the y-intercept is

(0, whatever you got)

How to find the x-intercepts:

Substitute 0 for y and solve for x:

Then the x-intercept is (whatever you got, 0)

Example 1:  

y = x² + 2x - 3

To find its y-intercept, substitute 0 for x, and solve for y

y = 0² + 2(0) - 3

y = -3

So the y-intercept is the point (0, -3)

To find its x-intercepts, substitute 0 for y, and solve for x

0 = x² + 2x - 3

x² + 2x - 3 = 0

(x + 3)(x - 1) = 0

x = -3, and x = 1

So the two x-intercepts are (-3, 0) and (1, 0)

Its graph looks like this

+graph%28+88%2C+88%2C+-5%2C+3%2C+-4%2C+4%2C+x%5E2%2B2x-3%29+ 

Example 2:

f(x) = -x² + x - 1

Replace f(x) by y

y = -x² + x - 1

To find its y-intercept, substitute 0 for x, and solve for y

y = -0² + 0 - 1

y = -1

So the y-intercept is the point (0, 1)

To find its x-intercepts, substitute 0 for y, and solve for x

0 = -x² + x - 1

x² - x + 1 = 0

That doesn't factor, so we use the
quadratic equation and find
           _
x = 1/2 ± Ö3/2i

which are imaginary, so there are no x-intercepts.

Its graph looks like this

+graph%28+88%2C+88%2C+-5%2C+3%2C+-4%2C+4%2C+-x%5E2%2Bx-1%29+ 

Example 3:  

y = x² + 6x + 9

To find its y-intercept, substitute 0 for x, and solve for y

y = 0² + 6(0) + 9

y = 9

So the y-intercept is the point (0, 9)

To find its x-intercepts, substitute 0 for y, and solve for x

0 = x² + 6x + 9

x² + 6x + 9 = 0

(x + 3)(x + 3) = 0

x = -3, and x = -3

These are the same so there is only one x-intercept
(-3, 0)

Its graph looks like this

+graph%28+90%2C+90%2C+-9%2C+2%2C+-1%2C+10%2C+x%5E2%2B6x%2B9%29+ 

It just "sits" on the x-axis at one point (-3, 0)

Edwin