Question

write the equation of the quadratic equation function i standard form represented by the graph

Answer 1

Answer:

$x^{2} +6x+8=0$

Step-by-step explanation:

we know that

The quadratic equation in standard form is equal to

$ax^{2} +bx+c=0$

In this problem we have

the vertex is equal to the point $(-3,-1)$

The zeros of the function are the points $(-4,0)$ and $(-2,0)$

The graph of the figure is a vertical parabola open upward

the equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2}+k$

where (h,k) is the vertex of the parabola

substitute

$y=a(x+3)^{2}-1$

with the point $(-2,0)$ find the value of a

$0=a(-2+3)^{2}-1$

$0=a(1)^{2}-1$

$a=1$

the equation in vertex form is equal to

$y=(x+3)^{2}-1$

Convert to standard form

$y=x^{2} +6x+9-1$

equate to zero

$x^{2} +6x+9-1=0$

$x^{2} +6x+8=0$