Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0). What is (a, b, c)?

xXxTenTacion Jul 16, 2018

#1**+2 **

Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0). What is (a, b, c)?

roots are \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

axis of symmetry is

\(x= \frac{-b}{2a}\\ \frac{-b}{2a}=3\\ -b=6a\\ b=-6a\\ \)

\(y=ax^2+bx+c\\ y=ax^2-6ax+c\\ sub \;\;in \;\;(1,0)\\ 0=a-6a+c\\ 0=c-5a\\ c=5a \)

\(y=ax^2-6ax+c\\ y=ax^2-6ax+5a\\ sub\;in\; (3,2)\\ 2=9a-18a+5a\\ 2=-4a\\ a=-\frac{1}{2}\\ \)

a=-0.5

c=-2.5

b=3

I do not have time to check so you had better do it :)

Melody Jul 16, 2018

#1**+2 **

Best Answer

Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0). What is (a, b, c)?

roots are \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

axis of symmetry is

\(x= \frac{-b}{2a}\\ \frac{-b}{2a}=3\\ -b=6a\\ b=-6a\\ \)

\(y=ax^2+bx+c\\ y=ax^2-6ax+c\\ sub \;\;in \;\;(1,0)\\ 0=a-6a+c\\ 0=c-5a\\ c=5a \)

\(y=ax^2-6ax+c\\ y=ax^2-6ax+5a\\ sub\;in\; (3,2)\\ 2=9a-18a+5a\\ 2=-4a\\ a=-\frac{1}{2}\\ \)

a=-0.5

c=-2.5

b=3

I do not have time to check so you had better do it :)

Melody Jul 16, 2018