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Solving Linear and Quadratic Equations and Absolute Value Equations

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 slope- intercept form of the equation for a
line. Another way to find the equation for a line is to use the
- 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:

ax2 + bx + c = 0

Then you can plug a, b, and c into the following 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:


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 + 0.17x2

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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