## anonymous 5 years ago Find the slope-predictor formula and write an equation for the line tangent to (a, f(a)) for f(x) = x^2 - 6x + 4. And can you please give me details of how you got the answer so that I ca uderstand it please?

1. anonymous

The slope-predictor function is also known as the derivative function. It is defined as$f'(x) = \lim_{\delta x \rightarrow 0}\frac{f(x+\delta x)-f(x)}{\delta x}$which comes from considering the slope of a secant between two points on a function, f(x), between x and x+delta x (I'm going through everything from the beginning since I only ever see 'slope-predictor' mentioned when definitions are used). Here,$f(x)=x^2-6x+4$and so by the definition,$f'(x)=\lim_{\delta x \rightarrow 0}\frac{[(x+\delta x)^2-6(x+\delta x)+4]-[x^2-6x+4]}{\delta x}$$=\lim_{\delta x \rightarrow 0}\frac{x^2+2x \delta x+(\delta x)^2-6x-6\delta x+4-x^2+6x-4}{\delta x}$$=\lim_{\delta x \rightarrow 0}\frac{(2x-6) \delta x+(\delta x)^2}{\delta x}$$=\lim_{\delta x \rightarrow 0}(2x-6+\delta x)$$=2x-6$

2. anonymous

This formula gives you the slope of the tangent to any point x on your function, f(x). So, at the point (a,f(a)), the slope is$m=2a-6$The equation of the line tangent to the function at this point is therefore given by the point-slope formula for a straight line:$y-y_1=m(x-x_1) \rightarrow y-f(a)=(2a-6)(x-a)$or$y=(2a-6)(x-a)+f(a)$

3. anonymous

You can use the fact that$f(a)=a^2-6a+4$to then write,$y=(2a-6)(x-a)+a^2-6a+4$and expand and simplify to obtain,$y=-a^2+2ax-6x+4$or better,$y=(2a-6)x+4-a^2$