+0

# help me

0
426
1

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)?

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 :)

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