First rewrite the quadratic function in standard form by completing the square, then find any x-intercepts and any y-intercepts.
-$$f ( x ) = x ^ { 2 } + 3 x + 2$$
A) $x$-intercepts: $- 1$ and $2 ; y$-intercept: 2
B) x-intercepts: $- 1$ and -2; y-intercept: 2
C) x-intercepts: 1 and 2; $y$-intercept: $- 1$
D) $x$-intercepts: 1 and -2; $y$-intercept: 0

Question

Quizplus · Accepted Answer

To rewrite the quadratic function in standard form by completing the square, we start with the given function $f(x) = x^2 + 3x + 2$. Completing the square involves creating a perfect square trinomial from the quadratic and linear terms:1. First, factor out the coefficient of $x^2$ if it is not 1 (in this case, it is already 1).2. Then, take half of the coefficient of $x$ (which is $3/2$) and square it ($(3/2)^2 = 9/4$).3. Add and subtract this square inside the function: $f(x) = (x^2 + 3x + 9/4) - 9/4 + 2$.4. This simplifies to $f(x) = (x + 3/2)^2 - 1/4$, which is the function in standard form.To find the x-intercepts, set $f(x) = 0$:$(x + 3/2)^2 - 1/4 = 0$$(x + 3/2)^2 = 1/4$Taking the square root of both sides gives $x + 3/2 = \pm 1/2$, leading to $x = -1$ and $x = -2$.The y-intercept is found by setting $x = 0$:$f(0) = 0^2 + 3(0) + 2 = 2$.Therefore, the x-intercepts are $-1$ and $-2$, and the y-intercept is $2$, which corresponds to choice B.

First Rewrite the Quadratic Function in Standard Form by Completing f(x)=x2+3x+2f ( x ) = x ^ { 2 } + 3 x + 2f(x)=x2+3x+2

First Rewrite the Quadratic Function in Standard Form by Completing $f ( x ) = x ^ { 2 } + 3 x + 2$