SOLUTION: Hi! I believe this is the correct category. For some reason my answers are not matching up. I am to graph the quadratic function, identify the axis of symmetry, and the coordinat

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hi! I believe this is the correct category. For some reason my answers are not matching up. I am to graph the quadratic function, identify the axis of symmetry, and the coordinat      Log On


   

Question 909603: Hi! I believe this is the correct category.
For some reason my answers are not matching up.
I am to graph the quadratic function, identify the axis of symmetry, and the coordinates of the vertex
For y=-x^2+5x-3
I got the axis of symmetry, -5/2
I got the y-intercept of (0,-3)
I got the x-intercept of 43/4
What am I doing wrong?
Thank you!
Found 4 solutions by jim_thompson5910, Alan3354, ewatrrr, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The axis of symmetry should be

x+=+-b%2F%282a%29

x+=+-5%2F%282%28-1%29%29

x+=+5%2F2 Notice how the two negatives cancel

This x value will be plugged back into y=-x%5E2%2B5x-3+ to find the y coordinate of the vertex.

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You have the correct y-intercept, but the x-intercept is incorrect.

You need to solve -x%5E2%2B5x-3+=0 for x.

You would use the quadratic formula to do this.

Answer by Alan3354(69443) About Me  (Show Source):
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
y=-x^2+5x-3 y-intercept(when x = 0) is -3
y = -(x-5/2)^2 + 25/4 - 12/4
y = -(x-5/2)^2 + 13/4
0= -(x-5/2)^2 + 13/4
x = 5/2 ± sqrt(13)/2
the vertex form of a Parabola opening up(a>0) or down(a<0), y=a%28x-h%29%5E2+%2Bk
where(h,k) is the vertex and x = h is the Line of Symmetry
axis of symmetry, x = 5/2
y-intercept, -3
x-intercepts: 5/2 ± sqrt(13)/2

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I am to graph the quadratic function, identify the axis of symmetry, and the coordinates of the vertex
For y = -x^2+5x-3
-x^2+5x = y+3
-(x^2-5x) = y+3
-(x^2 - 5x + (5/2)^2) = y+3 -(5/2)^2
-(x-(5/2))^2 = y+(12/4)-(25/4)
-(x-(5/2))^2 = y-(13/4)
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Axis of symmetry:: x = 5/2
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Vertex:: When x = 5/2, y = 13/4
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Cheers,
Stan H.
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