  # 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 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: 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: 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 + 0.17x2 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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