What is the tangent of-7/2 x^2+5x+5 ?

The tangent of-7/2 x^2+5x+5 is y=(-7a_{0}+5)x+7/2 a_{0}^2+5

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tangent of-7/2 x^2+5x+5

tangent of

- \frac{7}{2} x^{2} + 5 x + 5

y = (- 7 a_{0} + 5) x + \frac{7}{2} a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{7}{2} a_{0}^{2} + 5 a_{0} + 5)

Find the slope of

y = - \frac{7}{2} x^{2} + 5 x + 5 : \frac{d y}{d x} = - 7 x + 5

Compute the slope of

y = - \frac{7}{2} x^{2} + 5 x + 5

x = a_{0} : m = - 7 a_{0} + 5

Find the line with slope m=

- 7 a_{0} + 5

and passing through

(a_{0}, - \frac{7}{2} a_{0}^{2} + 5 a_{0} + 5) : y = (- 7 a_{0} + 5) x + \frac{7}{2} a_{0}^{2} + 5

y = (- 7 a_{0} + 5) x + \frac{7}{2} a_{0}^{2} + 5

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What is the tangent of-7/2 x^2+5x+5 ?
The tangent of-7/2 x^2+5x+5 is y=(-7a_{0}+5)x+7/2 a_{0}^2+5