Does anyone remember how to calculate the vertex of a quadratic equation?
I have to answer the following question:
Which is the vertex of the graph of the equation y = -x^2 + 2x + 3
help me please!
2007-01-04 10:21:25 · 5 answers · asked by mkn 2 in Science & Mathematics ➔ Mathematics
The equation for a parabola is:
y = a(x - h)² + k
Where the vertex is (h, k)
Transforming your equation into this form:
y = -x² + 2x + 3
y = -(x² - 2x) + 3
y = -(x² - 2x) - 1 + 1 + 3
y = -(x² - 2x + 1) + 1 + 3
y = -(x - 1)² + 4
Therefore the vertex is (1, 4), matching the h and k in the equation above.
2007-01-04 10:24:15 · answer #1 · answered by Jim Burnell 6 · 2⤊ 0⤋
Another way to do this is to get the axis of symmetry whicch is -b/2a = -2/-2= 1. therefore the x coordinate of the vertex is 1.
If you stick 1 into original equation you get y=-1+2+3 = 4
2007-01-04 18:30:48 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
The vertex formula:
y=(x-h)^2+k
y=-x^2+2x+3
y=-(x^2-2x)+3
y=-(x-2x+1)+3+1
y=-(x-1)^2+4
The vertex is (1,4)
2007-01-04 18:29:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋
-(x^2-2x+1)+4
y-4=-(x-1)^2
vertex=(1,4)
2007-01-04 18:24:56 · answer #4 · answered by raj 7 · 1⤊ 0⤋
x = -b+_ (Sqrt(b^2 - 4ac))/ 2a
2007-01-04 18:27:21 · answer #5 · answered by mud 1 · 1⤊ 0⤋