Consider the quadratic function f(x) = 2x2 – 8x – 10. The x-component of the vertex is . The y-component of the vertex is . The
2 answers:
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Compute the derivative:
![\dfrac{\mathrm d}{\mathrm dx}\bigg[8x^3+2x^2-8\bigg]=24x^2+4x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B8x%5E3%2B2x%5E2-8%5Cbigg%5D%3D24x%5E2%2B4x)
Set equal to zero and find the critical points:
Compute the second derivative at the critical points to determine concavity. If the second derivative is positive, the function is concave upward at that point, so the function attains a minimum at the critical point. If negative, the critical point is the site of a maximum.
![\dfrac{\mathrm d^2}{\mathrm dx^2}\bigg[8x^3+2x^2-8\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[24x^2+4x\bigg]=48x+4](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2%7D%7B%5Cmathrm%20dx%5E2%7D%5Cbigg%5B8x%5E3%2B2x%5E2-8%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B24x%5E2%2B4x%5Cbigg%5D%3D48x%2B4)
At
, the second derivative takes on the value of 
, so the function is concave upward, so the function has a minimum there of 
.
At
, the second derivative is 
, so the function is concave downward and has a maximum there of 
.
Set equal to zero and find the critical points:
Compute the second derivative at the critical points to determine concavity. If the second derivative is positive, the function is concave upward at that point, so the function attains a minimum at the critical point. If negative, the critical point is the site of a maximum.
At
At
The first equation is
The second equation is
If you sum the two equations you'll cancel y:
And since the sum must be 11, we have y=9.