What is the tangent of f(x)=x^2-3x+5,\at x=1 ?

The tangent of f(x)=x^2-3x+5,\at x=1 is y=-x+4

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tangent of f(x)=x^2-3x+5,\at x=1

tangent of

f (x) = x^{2} - 3 x + 5, at x = 1

y = - x + 4

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{2} - 3 x + 5 : \frac{df ( x )}{d x} = 2 x - 3

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 3) : y = - x + 4

y = - x + 4

n = 0 \sum \infty n^{3}

f (x) = - 7 x^{3} - 8 x^{2} - 5 x, at x = - 3

\int \frac{sin ( 2 x )}{1 + cos ( 2 x )} d x

\int_{1}^{2} e^{\frac{1}{x}} d x

\int (\frac{2 y}{( x ^{2} + y ^{2} ) ^{2}}) d y

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