Question

The zeros of a parabola are −5 and −3. The point (0, 60) is on the graph as represented by the equation. 60 = a(0 + 5)(0 + 3) What is the value of a?

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Zeros of the parabola are -5 and -3}$

$\underline{\textsf{To find:}}$

$\textsf{The value of a}$

$\mathsf{}$

$\underline{\textsf{Solution:}}$

$\textsf{Since the zeros of the parabola are -5 and -3, we can write}$

$\textsf{the equation of the parabola as}$

$\mathsf{y=k(x+5)(x+3)}$

$\mathsf{y=k(x^2+8x+15)}$

$\textsf{It passes through (0,60)}$

$\mathsf{60=k(0+8(0)+15)}$

$\mathsf{60=15k}$

$\implies\mathsf{k=4}$

$\textsf{The equation of the parabola becomes}$

$\mathsf{y=4(x^2+8x+15)}$

$\mathsf{y=4(x^2+8x)+60}$

$\mathsf{y-60=4(x^2+8x)}$

$\textsf{To make R.H.S of the equation as a perfect square}$

$\mathsf{y-60=4((x^2+8x+16)-16)}$

$\mathsf{y-60=4(x+4)^2-64}$

$\mathsf{4(x+4)^2=y+4}$

$\mathsf{(x+4)^2=\dfrac{1}{4}(y+4)}$

$\textsf{Comparing this with}\;\mathsf{(x-h)^2=4a(y-k)}\,\textsf{we get}$

$\mathsf{4a=\dfrac{1}{4}}$

$\implies\boxed{\mathsf{a=\dfrac{1}{16}}}$

$\underline{\textsf{Answer:}}$

$\textsf{The value of a is}\;\mathsf{\dfrac{1}{16}}$