What is the tangent of y=8sqrt(x)-6,(64,58) ?

The tangent of y=8sqrt(x)-6,(64,58) is y= 1/2 x+26

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tangent of y=8sqrt(x)-6,(64,58)

tangent of

y = 8 x - 6, (64, 58)

y = \frac{1}{2} x + 26

Solution steps

Find the slope of

y = 8 x - 6 : \frac{d y}{d x} = \frac{4}{x}

Compute the slope of

y = 8 x - 6

(64, 58) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(64, 58) : y = \frac{1}{2} x + 26

y = \frac{1}{2} x + 26

\int_{7}^{9} (x^{2} + 2 x - 5) d x

\int x^{2} e^{4 x} d x

\frac{1}{s ^{2} + 6 s + 25}

f (x) = \frac{7 x}{e ^{x}}

\int \frac{6}{y ^{2} - 2 y - 8} d y

What is the tangent of y=8sqrt(x)-6,(64,58) ?
The tangent of y=8sqrt(x)-6,(64,58) is y= 1/2 x+26