Question

Which of the following gives the correct intercept points and vertex point for the function f(x) = x2 + 5x + 6?

Answer 1

We are given the following quadratic equation.

$f(x)=x^2+5x+6$

Recall that the standard form of a quadratic equation is given by

$f(x)=ax^2+bx+c$

Comparing the given equation with the standard form, the coefficients are

a = 1

b = 5

c = 6

The vertex point is given by

$h=-\frac{b}{2a}=-\frac{5}{2(1)}=-\frac{5}{2}$$\begin{gathered} f(-\frac{5}{2})=(-\frac{5}{2})^2+5(-\frac{5}{2})+6 \\ f(-\frac{5}{2})=\frac{25}{4}^{}-\frac{25}{2}+6 \\ f(-\frac{5}{2})=-\frac{1}{4} \end{gathered}$

So, the vertex point is

$vertex\: point=(-\frac{5}{2},-\frac{1}{4})$

Now let us find the y-intercept of this equation.

Substitute x = 0 into the equation

$\begin{gathered} f(x)=x^2+5x+6 \\ f(0)=0^2+5(0)+6 \\ f(0)=6 \end{gathered}$

So, the y-intercept is

$y-intercept=(0,6)$

Finally, let us find the x-intercept of this equation

Substitute f(x) = 0 into the equation

$\begin{gathered} 0=x^2+5x+6 \\ 0=x^2+3x+2x+6 \\ 0=x(x+3)+2(x+3)_{} \\ 0=(x+2)\mleft(x+3\mright)_{} \\ (x+2)=0\: \: and\: \: (x+3)=0\: _{} \\ x=-2\: \: and\: \: x=-3 \end{gathered}$

So, the x-intercepts are

$x-intercepts=(-2,0)\: and\: (-3,0)_{}$

Conclusion:

$y-intercept=(0,6)\quad vertex\: point=(-\frac{5}{2},-\frac{1}{4})\quad x-intercepts=(-2,0)\: and\: (-3,0)_{}$

Therefore, the correct answer is option B