Answer
$y = -\frac{2}{3}x^2+\frac{14}{3}x$
Work Step by Step
$y = ax^2+bx+c$
$y' = 2ax+b$
When $x = -1,~~$ then $~~y' = 6$:
$y' = 2a(-1)+b = 6$
$b = 2a+6$
When $x = 5,~~$ then $~~y' = -2$:
$y' = 2a(5)+b = -2$
$10a+(2a+6) = -2$
$12a = -8$
$a = -\frac{2}{3}$
Then:
$b = 2a+6 = 2(-\frac{2}{3})+6 = \frac{14}{3}$
When $x = 1,~~$ then $~~y = 4$:
$y = ax^2+bx+c$
$(-\frac{2}{3})(1)^2+(\frac{14}{3})(1)+c = 4$
$-\frac{2}{3}+\frac{14}{3}+c = 4$
$c = 4-4$
$c = 0$
We can write the equation of the parabola:
$y = -\frac{2}{3}x^2+\frac{14}{3}x$
