Find the coordinates of the turning point of the curve y=8x+(1/2x^2) and determine whether this point is a maximum or minimum point?

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Answer 1 · 2018-01-18T22:29:46

$(- 8 . - 32)$ $(-8.-32)$ ; minimum point

$y = 8 x + (\frac{1}{2} x^{2})$ $y = 8x + (1/2x^2)$

$f (x) = \frac{1}{2} x^{2} + 8 x$ $f(x) = 1/2x^2 + 8x$

$(Deltay)/(Deltax) (x^2/2) = 2 * x/2 = x$

$(Deltay)/(Deltax) (8x) = 8$

$(Deltay)/(Deltax) (1/2x^2 + 8x) = x+8$

at a turning point, $(Deltay)/(Deltax) = 0$ .

$(Deltay)/(Deltax) = x+8 = 0$

$x = 0-8 = -8$

either side of $x=-8:$

when $x= -7.9999$ , $(Deltay)/(Deltax) = 0.0001$

when $x = -8$ , $(Deltay)/(Deltax) = 0$

when $x = -8.0001$ , $(Deltay)/(Deltax) = -0.0001$

to the left of $x=-8$ , the gradient is negative.
at $x-8$ , the gradient is $0$ .
to the right of $x=-8$ , the gradient is positive.

this shows that $(-8, y)$ is a minimum point.

graph{8x + (1/2)x^2 [-80, 80, -40, 40]}

$y = 8x + (1/2)x^2$

$y = -64 + (64/2)$

$=-64 + 32$

$=-32$

$f(-8) = -32)$