What is the tangent of f(x)=4x^2-5x,(-1,9) ?

The tangent of f(x)=4x^2-5x,(-1,9) is y=-13x-4

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tangent of f(x)=4x^2-5x,(-1,9)

tangent of

f (x) = 4 x^{2} - 5 x, (- 1, 9)

y = - 13 x - 4

Solution steps

Find the slope of

f (x) = 4 x^{2} - 5 x : \frac{df ( x )}{d x} = 8 x - 5

Compute the slope of

f (x) = 4 x^{2} - 5 x

(- 1, 9) : m = - 13

Find the line with slope m=

- 13

and passing through

(- 1, 9) : y = - 13 x - 4

y = - 13 x - 4

n = 1 \sum \infty \frac{3 ^{n}}{n ^{3}}

\int_{0}^{1} 7 x x^{2} + 9 d x

\int \frac{x ^{2} + 1}{( x - 7 ) ( x - 6 )} d x

n = 1 \sum \infty x^{2 n}

h \to 0 lim (\frac{( 2. 8 ^{h} - 1 )}{h})

What is the tangent of f(x)=4x^2-5x,(-1,9) ?
The tangent of f(x)=4x^2-5x,(-1,9) is y=-13x-4