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anonymous
 5 years ago
Find the slopepredictor 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?
anonymous
 5 years ago
Find the slopepredictor 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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The slopepredictor 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 'slopepredictor' mentioned when definitions are used). Here,\[f(x)=x^26x+4\]and so by the definition,\[f'(x)=\lim_{\delta x \rightarrow 0}\frac{[(x+\delta x)^26(x+\delta x)+4][x^26x+4]}{\delta x}\]\[=\lim_{\delta x \rightarrow 0}\frac{x^2+2x \delta x+(\delta x)^26x6\delta x+4x^2+6x4}{\delta x}\]\[=\lim_{\delta x \rightarrow 0}\frac{(2x6) \delta x+(\delta x)^2}{\delta x}\]\[=\lim_{\delta x \rightarrow 0}(2x6+\delta x)\]\[=2x6\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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\[m=2a6\]The equation of the line tangent to the function at this point is therefore given by the pointslope formula for a straight line:\[yy_1=m(xx_1) \rightarrow yf(a)=(2a6)(xa)\]or\[y=(2a6)(xa)+f(a)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the fact that\[f(a)=a^26a+4\]to then write,\[y=(2a6)(xa)+a^26a+4\]and expand and simplify to obtain,\[y=a^2+2ax6x+4\]or better,\[y=(2a6)x+4a^2\]
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