What is the tangent of y=-0.02x^2-0.6x+90 ?

The tangent of y=-0.02x^2-0.6x+90 is y=(-0.04a_{0}-0.6)x+0.02a_{0}^2+90

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tangent of y=-0.02x^2-0.6x+90

tangent of

y = - 0.02 x^{2} - 0.6 x + 90

y = (- 0.04 a_{0} - 0.6) x + 0.02 a_{0}^{2} + 90

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 0.02 a_{0}^{2} - 0.6 a_{0} + 90)

Find the slope of

y = - 0.02 x^{2} - 0.6 x + 90 : \frac{d y}{d x} = - 0.04 x - 0.6

Compute the slope of

y = - 0.02 x^{2} - 0.6 x + 90

x = a_{0} : m = - 0.04 a_{0} - 0.6

Find the line with slope m=

- 0.04 a_{0} - 0.6

and passing through

(a_{0}, - 0.02 a_{0}^{2} - 0.6 a_{0} + 90) : y = (- 0.04 a_{0} - 0.6) x + 0.02 a_{0}^{2} + 90

y = (- 0.04 a_{0} - 0.6) x + 0.02 a_{0}^{2} + 90

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What is the tangent of y=-0.02x^2-0.6x+90 ?
The tangent of y=-0.02x^2-0.6x+90 is y=(-0.04a_{0}-0.6)x+0.02a_{0}^2+90