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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?

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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,\[y=(2a-6)x+4-a^2\]

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