Solving Linear and Quadratic Equations and Absolute Value Equations
Slide 11 
Example: Find the Equation for a Line
3. Substitute the x and y values for one of the points (2,6) and
solve for b
6 = 2(2) + b
6 = 4+b
6 −4 = b => b = 2
y = 2x + 2
4. Check your answer using the x and y values for the other point
0 = 2(−1) + 2? Yes! 
Slide 12 
Example: Find the Equation for a Line
y = 2x + 2

Slide 13 
Another Way to Find the Equation for a Line
We have been using the slopeintercept form of the equation for a
line. Another way to find the equation for a line is to use the
point  slope method .
m = slope
= one point on the line
So, given slope = 2 and point
= (−1, 0):
y − 0 = 2(x − (−1))
y = 2x + 2
Same as before! 
Slide 14 
Solving Quadratic Equations
You may need to find the solution to a quadratic equation. To do
this, you must use the distributive , additive , and multiplicative
properties to get the equation into this form:
ax^{2} + bx + c = 0
Then you can plug a, b, and c into the fol lowing equation , which is
called the quadratic formula .
is called the
discriminant. 
Slide 15 
Solving Quadratic Equations
The solution to a quadratic equation specifies where it crosses the x
axis. A quadratic equation may have 2 solutions:
A quadratic equation may have no solutions:

Slide 16 
Solving Quadratic Equations
A quadratic equation may have one solution:

Slide 17 
Solving Quadratic Equations: Example
So a = 4, b = 12 and c = 6. 
Slide 18 
Solving Quadratic Equations: Example
a = 4, b = 12, c = 6
The two solutions are 0.634 and 2.366. 
Slide 19 
Solving Quadratic Equations: Example
Check the solutions:
Good! 
Slide 20 
Solving Quadratic Equations: Example
We can graph quadratic equations in a manner similar to that for
linear functions :

Slide 21 
Graphing Quadratic Equations

Slide 22 
Solving Quadratic Equations: Example
Recall the 8 animals who received different doses of a drug and
whose weight gain was measured. The quadratic equation that best
described the relationship between dose and weight gain was:
y = 1.13 − 0.41x + 0.17x^{2
}
We can use substitution to find the predicted weight gain, given a
dose. For example, if we know an animal like these received dose 3,
we would predict that the weight gain would be
1.13 − 0.41(3) + 0.17(3)(3) = 1.43 dekagrams.
What if we knew the animal had gained 5 dekagrams, and wanted
to deduce what the dose had been? 
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