SOLUTION: How would I find the x and y-intercept and vertex of: H(x)=3x-1/2x^2 Thank you

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Question 948400: How would I find the x and y-intercept and vertex of: H(x)=3x-1/2x^2
Thank you
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
H(x)=3x-1/2x^2 Write the equation in the form:
ax%5E2%2Bbx%2Bc=0
In this case:
%28-1%2F2%29x%5E2%2B3x=y
The x-intercept is where y=0 so
%28-1%2F2%29x%5E2%2B3x%2B0=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -0.5x%5E2%2B3x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%283%29%5E2-4%2A-0.5%2A0=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+9+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+9+%29%29%2F2%5C-0.5+=+0
x%5B2%5D+=+%28-%283%29-sqrt%28+9+%29%29%2F2%5C-0.5+=+6

Quadratic expression -0.5x%5E2%2B3x%2B0 can be factored:
-0.5x%5E2%2B3x%2B0+=+-0.5%28x-0%29%2A%28x-6%29
Again, the answer is: 0, 6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-0.5%2Ax%5E2%2B3%2Ax%2B0+%29

The answers are 0 and 6, so the x-intercepts are (0,0) and (6,0)
And the y-intercept(s) is where x=0, so:
y=%28-1%2F2%290%5E2%2B3%280%29
y=0 So the y-intercept is at the origin (0,0)
The axis of symmetry is given by: (This is the x value of the vertex)
x=-b%2F2a
In this case:
x=%28-3%29%2F%282%28-1%2F2%29%29
x=3 Substitute this in the original equation to get the y value of the vertex.
y=%28-1%2F2%293%5E2%2B3%283%29
y=%28-1%2F2%29%289%29%2B9
y=-4.5%2B9
y=4.5 So the vertex is (3,4.5)