Answer:
[tex]y=2x^2+4x[/tex]
Step-by-step explanation:
Given table:
[tex]\begin{array}{|c|c|c|c|c|}\cline{1-5} x & -3 & -2 & -1 & 0\\\cline{1-5} f(x) & 6 & 0 & -2 & 0\\\cline{1-5}\end{array}[/tex]
The x-intercepts of a quadratic function are when f(x) = 0.
Therefore, the x-intercepts are: x = -2 and x = 0.
Intercept form of a quadratic equation
[tex]y=a(x-p)(x-q)[/tex]
where:
Substitute the found x-intercepts and one of the points from the table into the formula and solve for a:
[tex]\begin{aligned} y&=a(x-p)(x-q)\\\\\implies6&=a(-3-(-2))(-3-0)\\6&=a(-1)(-3)\\6&=3a\\a&=\dfrac{6}{3}\\\implies a&=2\end{aligned}[/tex]
Substitute the x-intercepts and the found value of a into the formula:
[tex]\implies y=2(x+2)(x-0)[/tex]
Expand to standard form:
[tex]\implies y=2(x+2)(x-0)[/tex]
[tex]\implies y=2x(x+2)[/tex]
[tex]\implies y=2x^2+4x[/tex]