SOLUTION: What is the vertex of the quadratic formula y=-x^2+3x+4? I'm trying to do it without a calculator so I can get the vertex every time I try to do a quadratic equation. I've already

Question 976873: What is the vertex of the quadratic formula y=-x^2+3x+4? I'm trying to do it without a calculator so I can get the vertex every time I try to do a quadratic equation. I've already got the line of symmetry, the y-intercepts, the x-intercepts, the graph, the minimum, the maximum, and the points and I've plotted everything, but two points don't make the vertex a point with whole numbers. If you have a way of getting the vertex of the graph without using a calculator, please let me know.
Found 2 solutions by josgarithmetic, addingup:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Yes, a lesson is available for how to change a quadratic equation of y=degreeTwoExpression into standard form, also known as vertex form.

Study this:
LESSON: How to solve AND how to Complete The Square for quadratic equation
http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
y= -x^2+3x+4 is in the quadratic form y= ax2 + bx + c
The vertex (h, k) is found by computing h = �b/2a, and then evaluating y at h to find k.
--------------------------------------
h = �b/2a = �(3)/2(1) = �3/2 or -1 1/2, this is one of our vertex points.
Now we can find k by evaluating y at h= -3/2:
k= (-3/2)^2+3(-3/2)+4
= 9/4 -9/2 + 4
=9/4 - 18/4 + 16/4
= (9-18+16)/4
= 7/4 or 1 3/4
The vertex is at (-1 1/2, 1 3/4)

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