Question

The zeros of a parabola are –4 and 2, and (6, 10) is a point on the graph. Which equation can be solved to determine the value of a in the equation of the parabola? 6 = a(10 – 4)(10 + 2) 6 = a(10 + 4)(10 – 2) 10 = a(6 – 4)(6 + 2) 10 = a(6 + 4)(6 – 2)

Answer 1

the answer is D

Explanation:

Answer 2

`\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------\\\\ \stackrel{\textit{zeros\qquad \qquad \qquad }}{ \begin{cases} x=-4\implies &x+4=0\\ x=2\implies &x-2=0 \end{cases}} \\\\\\ \stackrel{\textit{factored vertex form}}{y=a(x+4)(x-2)}\qquad (6,10)\textit{ we also know that } \begin{cases} x=6\\ y=10 \end{cases} \\\\\\ 10=a(6+4)(6-2)`