SOLUTION: how do you write y=(x-4)(x+2) in standard form? with the standard form of the equation use the quadratic formula to identify the x-value of the vertex. use the quadratic formula w

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: how do you write y=(x-4)(x+2) in standard form? with the standard form of the equation use the quadratic formula to identify the x-value of the vertex. use the quadratic formula w      Log On


   

Question 1024230: how do you write y=(x-4)(x+2) in standard form?
with the standard form of the equation use the quadratic formula to identify the x-value of the vertex. use the quadratic formula with the vertex -b/2a then substitute the x-value of the vertex into the original equation to find the y-value of the vertex then write the coordinates of the vertex. what are the 3 points to graph an parabola? PLEASE HELP
Found 3 solutions by josgarithmetic, MathTherapy, robertb:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
The term -b%2F%282a%29 is based on ax%5E2%2Bbx%2Bc=y.

You may be able to do the problem on your own.
Perform the multiplication on the right side member.
Identify the values of a and b.
Form the term needed for -b%2F%282a%29; and use this to find the value of y.

You may also benefit from a helpful instructive video:
changing quadratic equation from general to standard form

Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!

how do you write y=(x-4)(x+2) in standard form?
with the standard form of the equation use the quadratic formula to identify the x-value of the vertex. use the quadratic formula with the vertex -b/2a then substitute the x-value of the vertex into the original equation to find the y-value of the vertex then write the coordinates of the vertex. what are the 3 points to graph an parabola? PLEASE HELP
Standard form of equation: y = (x - 4)(x + 2): highlight_green%280+=+x%5E2+-+2x+-+8%29

The x-value of the vertex is: x+=+-+b%2F%282a%29 ----> x+=+-+-+2%2F%282+%2A+1%29%29%7D%7D+------%3E+%7B%7B%7Bx+=+2%2F2 ------> highlight_green%28x+=+1%29

y+=+x%5E2+-+2x+-+8

y+=+%281%29%5E2+-+2%281%29+-+8  ------ Substituting 1 for x

y = 1 - 2 - 8 -------> highlight_green%28y+=+-+9%29 ------ y-value of vertex

Coordinates of vertex: highlight_green%28matrix%281%2C3%2C+%22%281%22%2C+%22%2C%22%2C+%22-+9%29%22%29%29



(x - 4) = 0        and       (x + 2) = 0

Roots: matrix%281%2C3%2C+x+=+4%2C+AND%2C+x+=+-+2%29



3 points to graph this parabola:

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
You don't even have to use formulas to find the vertex of the parabola if you know the roots.
y = (x-4)(x+2) ==> the roots of the equation are 4 and -2. The x-coordinate of the vertex should be the average of the two roots, namely %284%2B%28-2%29%29%2F2+=+1. To find the y-coordinate of the vertex plug it into the original equation:
y = (1-4)(1+2) = -3*3 = -9.
==> (h,k) = (1,-9)

==> y+=+x%5E2+-+2x+-+8, the standard form of the equation