First rewrite the quadratic function in standard form by completing the square, then find any x-intercepts and any y-intercepts.
-$$f ( x ) = - 9 x ^ { 2 } - 2 x - 12$$
A) No x-intercepts; No $y$-intercepts
B) x-intercepts: $- \frac { 1 } { 9 }$ and $- \frac { 107 } { 9 } ; y$-intercept: 12
C) $x$-intercepts: $\frac { 1 } { 9 }$ and $\frac { 107 } { 9 } ; y$-intercept: $- 12$
D) No $x$-intercepts; $y$-intercept: $- 12$

Question

Quizplus · Accepted Answer

To complete the square and rewrite the function in standard form, we first factor out the coefficient of $x^2$ from the first two terms: $$f(x) = -9(x^2 + \frac{2}{9}x) - 12.$$ Completing the square within the parentheses involves adding and subtracting $(\frac{1}{9})^2 = \frac{1}{81}$ inside the parentheses, but since we factored out -9, we actually add $9 \times \frac{1}{81} = \frac{1}{9}$ to maintain equality. This gives us $$f(x) = -9(x^2 + \frac{2}{9}x + \frac{1}{81}) - 12 - \frac{1}{9} = -9(x + \frac{1}{9})^2 - \frac{109}{9}.$$ The vertex form reveals that the parabola opens downwards (since the coefficient of the squared term is negative) and has its vertex at $(- \frac{1}{9}, - \frac{109}{9})$. Since the vertex's y-coordinate is below the x-axis and the parabola opens downwards, there are no x-intercepts. The y-intercept occurs when $x = 0$, so substituting $x = 0$ into the original equation gives $f(0) = -9(0)^2 - 2(0) - 12 = -12$. Therefore, there are no x-intercepts, and the y-intercept is -12.

First Rewrite the Quadratic Function in Standard Form by Completing f(x)=−9x2−2x−12f ( x ) = - 9 x ^ { 2 } - 2 x - 12f(x)=−9x2−2x−12

First Rewrite the Quadratic Function in Standard Form by Completing $f ( x ) = - 9 x ^ { 2 } - 2 x - 12$