I'm having a problem finding a vertex point!! please help!! explain?

Question

hi i am in calculus and i need to find the vertext point to y=x^2-4x+7 i need some to explain how to do it!! PLEASE!!!

Answer 1

First, take the derivative.

dy/dx = 2x - 4

Set the derivative to a slope of zero and solve.

0 = 2x - 4

x=2

This is the x-coordinate of the vertex. Sub it in the equation to get the y-coordinate.

y=2^2-4(2)+7=3

The vertex point is at (2,3)

Answer 2

A vertex is a place on a curve where the slope is zero (horizontal line; in y = mx+b, m = 0, so y = b everywhere no matter what 'x' is). It doesn't matter if the curve of the line is "dipping down then up", or "over the top", the place where it changes slopes from pos to neg, or neg to pos, it has to go through a zero slope.

Since you should know how to find the slope of a line (differentiation), and you know the slope is 'zero' at the vertex, you should be able to find x, then y (the answers above have worked it out, nicely).

.

Answer 3

The x value of the vertex is -b/2a= 4/2*1 = 2
Now find y=f(2) = 2^2 -4*2+7 = 3.
So vertex is (2,3)

Answer 4

y=x^2-4x+7=(x-2)^2+3
Vertex: (2, 3)
------
Ideas: Compare to the vertex form y = (x-h)^2+k, where (h,k) is the coordinates of the vertex.

Answer 5

Take the derivative: dy/dx

Set dy/dx = 0 and solve for 'x'.

Then plug this back into your original equation and find the corresponding 'y' value. That will be the coordinate for the vertex.