SOLUTION: find the vertex and intercept coordinates of the following. y=x^2-5x-4

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Question 143459: find the vertex and intercept coordinates of the following.
y=x^2-5x-4
Answer by nabla(475) About Me  (Show Source):
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Set y=0 :
x^2-5x-4=0
Use the quadratic formula to find where y is zero (the x-intercepts):
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-4+=+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=%28-5%29%5E2-4%2A1%2A-4=41.

Discriminant d=41 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+41+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+41+%29%29%2F2%5C1+=+5.70156211871642
x%5B2%5D+=+%28-%28-5%29-sqrt%28+41+%29%29%2F2%5C1+=+-0.701562118716424

Quadratic expression 1x%5E2%2B-5x%2B-4 can be factored:
1x%5E2%2B-5x%2B-4+=+1%28x-5.70156211871642%29%2A%28x--0.701562118716424%29
Again, the answer is: 5.70156211871642, -0.701562118716424. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-4+%29


To find the y intercept we set x=0;
0^2-5(0)-4=y
It follows that y=-4. Thus, (0,-4) is the y-intercept.
The vertex is x=-b/(2a) when the equation is in form ax^2+bx+c=0. Thus, 5/2=x is the vertex. Now, find the corresponding y coordinate for x=5/2.
(5/2)^2-5(5/2)-4=25/4-50/4-16/4=-41/4
Thus, the vertex is at (5/2,-41/4). The graph is above.