What is the tangent of y=6x^2-7x+4,(1,3) ?

The tangent of y=6x^2-7x+4,(1,3) is y=5x-2

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tangent of y=6x^2-7x+4,(1,3)

tangent of

y = 6 x^{2} - 7 x + 4, (1, 3)

y = 5 x - 2

Solution steps

Find the slope of

y = 6 x^{2} - 7 x + 4 : \frac{d y}{d x} = 12 x - 7

Compute the slope of

y = 6 x^{2} - 7 x + 4

(1, 3) : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

sin (3 x) cos (2 x)

\int sin^{4} (2 x) cos^{2} (2 x) d x

\int 3 sin (5 x) d x

\int \frac{1}{1 + x ^{\frac{3}{2}}} d x

\frac{d}{d t} (\frac{tan ( t )}{cos ( t ) - 4})

What is the tangent of y=6x^2-7x+4,(1,3) ?
The tangent of y=6x^2-7x+4,(1,3) is y=5x-2