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We are given the following quadratic equation

The vertex is the maximum/minimum point of the quadratic equation.
The x-coordinate of the vertex is given by

Comparing the given equation with the general form of the quadratic equation, the coefficients are
a = 2
b = 7
c = -10

The y-coordinate of the vertex is given by

This means that we have a minimum point.
Therefore, the minimum point of the given quadratic equation is
-3x-6y=17
-6y=3x+17
y=-x/2-17/6 so the slope of the line is -1/2
For another line to be perpendicular to this one the product of their slopes must equal -1
(-1/2)m=-1, m=2 so the perpendicular line will have a slope of 2...
y=2x+b, using point (6,3) we can solve for b...
3=2(6)+b
3=12+b
b=-9 so
y=2x-9
(3x^5 + 8x^3) - (7x^2 - 6x^3) = 3x^5 + 8x^3 - 7x^2 - 6x^3 = 3x^5 + 2x^3 - 7x^2