MHB Find the equation of the tangent

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To find the point where F'(a) = a for the function f(x) = -x^2 + 3x - 7, first calculate the derivative, yielding f'(x) = -2x + 3. Setting f'(a) equal to a results in the equation a = -2a + 3, which simplifies to a = 1. Evaluating f at this point gives f(1) = -5. The equation of the tangent line at the point (1, -5) is y = x - 6.
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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
 
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shawen said:
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., $y = x - 6$.
 
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