# Page 43 - 1 = ax 2-2a)x 6 We know that c is the y-intercept...

This preview shows page 1. Sign up to view the full content.

Huynh, Kazyak, Rees, Ufner Page 43, #22 Given information: f(x) = ax 2 + bx + c kkkkkkkkkkkkkkk (1,1) is on the graph of f(x) k (1,1) is the vertex of f(x) k The axis of symmetry is x = -b/2a k The y-intercept of the graph is (0,6) With this information, we are asked to find a quadratic function to fit all given information. We plot the points (1,1) and (0,6) on the graph. Because this is a positive porabola, we know that (2,6) will also be a point on the graph. Based on our graph, the axis of symmetry is x=1. Therefore, 1=-b/2a. We decided to solve for b, so we can plug it into the equation. 1=-b/2a ……. -2a = b
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 = ax 2 + (-2a)x + 6 We know that c is the y-intercept, which equals 6. 1 = a – 2a + 6 We also know that a must be positive, since the porabola opens -5 = -a up. We set the equation equal to 1 because we used the point a = 5 (1,1). 1 replaces all x’s and f(x). Now that we know what a kkkk equals, we can find b. -2a = b (-2)(5) = b b = -10 We now have solved for a, b, and c. We can conclude that our k quadratic function is f(x) = 5x 2- 10x + 6. To check our work, k we plugged the equation into our calculators. Our graph was kk consistent with all of the information given to us. Jon—graph here...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern