Question

Which equation represents a graph with a vertex at (–3, 2)?

Answer 1

Y = 4x² + 24x +38

I hope this helps. (:

Answer 2

We will proceed to find the equations in the vertex form in each case

Part a) $y= 4x^{2}+24x+38$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-38= 4x^{2}+24x$

Factor the leading coefficient

$y-38= 4(x^{2}+6x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y-38+36= 4(x^{2}+6x+9)$

$y-2= 4(x+3)^{2}$

$y= 4(x+3)^{2}+2$

the vertex is the point $(-3,2)$

Part b)  $y=4x^{2}-24x+38$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-38= 4x^{2}-24x$

Factor the leading coefficient

$y-38= 4(x^{2}-6x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y-38+36= 4(x^{2}-6x+9)$

$y-2= 4(x-3)^{2}$

$y= 4(x-3)^{2}+2$

the vertex is the point $(3,2)$

Part c) $y=4x^{2}+12x+2$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-2= 4x^{2}+12x$

Factor the leading coefficient

$y-2= 4(x^{2}+3x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y-2+9= 4(x^{2}+3x+2.25)$

$y+7= 4(x+1.5)^{2}$

$y= 4(x+1.5)^{2}-7$

the vertex is the point $(-1.5,7)$

Part d) $y=4x^{2}+16x+13$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-13= 4x^{2}+16x$

Factor the leading coefficient

$y-13= 4(x^{2}+4x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y-13+16= 4(x^{2}+4x+4)$

$y+3= 4(x+2)^{2}$

$y= 4(x+2)^{2}-3$

the vertex is the point $(-2,-3)$

therefore

the answer is

$y= 4x^{2}+24x+38$