+586 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)?
+118687 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 :)
+118687 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 :)
