What is the vertex form of the equation? y=x^2+4x-3?

y = (x + 2)^2 - 7

Step-by-step explanation:

The vertex form of the equation of a parabola (such as we have here) is y = a(x-h)^2 + k, where (h,k) is the vertex.  It's safe to assume that the coefficient "a" in y=x^2+4x-3 is 1.

The best thing to do now is to complete the square.

That is, y = x^2 + 4x  + 4 - 4                                -3

This can be re-written in more compact form as:

            y = (x + 2)^2 - 7

Comparing this to

            y = a(x-h)^2 + k

shows that h = -2 and k = -7, which leads to the conclusion that the vertex is at (-2, -7).

understand that i don't know the answers

will someone answer his question

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9