Asked by Renee A
How do you go about finding the roots\x intercepts of the function; 2x^2-16x+36 so that I can graph it ?
Thank you !
Thank you !
Answers
Answered by
Steve
First, divide by 2.
x^2-8x+18 = 0
the roots are provided by the quadratic formula:
x = (8±√(8^2-(4*1*18))/2
You can see that they are complex because the discriminant is negative. So, there are no x-intercepts.
The y-intercept is of course 36, since there, x=0.
To find the vertex, complete the square.
y = 2x^2-16x+26
= 2(x^2-8x) + 36
= 2(x^2-8x+16) + 36 - 2*16
= 2(x-4)^2 + 4
So, the vertex is at (2,4)
Now you should be able to sketch it, using symmetry around the line x=2.
x^2-8x+18 = 0
the roots are provided by the quadratic formula:
x = (8±√(8^2-(4*1*18))/2
You can see that they are complex because the discriminant is negative. So, there are no x-intercepts.
The y-intercept is of course 36, since there, x=0.
To find the vertex, complete the square.
y = 2x^2-16x+26
= 2(x^2-8x) + 36
= 2(x^2-8x+16) + 36 - 2*16
= 2(x-4)^2 + 4
So, the vertex is at (2,4)
Now you should be able to sketch it, using symmetry around the line x=2.
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