$$f(x)$$ is a quadratic function whose graph has a vertex at the point $$(-3 , 2)$$ and has a y-intercept at the point $$(0 , -16)$$.

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Uniteasy Admin · Accepted Answer

Since the vertex of the quadratic function is at the point $$ (-3 , 2)$$,
The equation for function is
$$ f(x)=a(x+3)^2+2 $$
Since the function has a y-intercept at the point $$(0 , -16)$$, we get the following equation
$$-16 = a(0+3)^2+2$$
Solve the equation
$$a = -2$$
Therefore, the  equation for quadratic function is
$$ f(x)=-2(x+3)^2+2 $$
 If $$f(x) = 0$$,
$$-2(x+3)^2+2 = 0$$
(x+3)^2=1
$$x+3=\pm 1$$
$$x=-2$$ and $$x=-4$$
Therefore, the x-intercepts of the graph of f(x) are points $$(-4,0) $$ and $$(-2,0)$$

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