Find the equation of the tangent

[shawen]

determine the point (a,F(a)) for which F'(a)=a, given that f(x)= -x^2+3x-7. write the equation of the tangent to f(x) at the point found

1) i have tried putting into first principle but when i do my denominator become 0 so i am kind stuck , i don't know what to do

please help me
thanks so much

 

[Euge]

determine the point (a,F(a)) for which F'(a)=a, given that f(x)= -x^2+3x-7. write the equation of the tangent to f(x) at the point found

Hi shawen,

First compute $f'(x) = -2x + 3$. Since $f'(a) = a$, we have $a = -2a + 3$, or $a = 1$. Then $f(a) = f(1) = -5$. The equation of the tangent to $f$ at $(1,-5)$ is $y = -5 + f'(1)(x - 1)$, or $y = -5 + 1 (x - 1)$, i.e., $y = x - 6$.