The diagram shows the line $2y=x+5$ and the curve $y=x^2-4x+7$, which intersects the points A and B. How do you find the $x$-coordinates and also the equation of the tangent to the curve at B?
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$2y=x+5$ implies $y=\frac{x}{2}+\frac{5}{2}$
Let $y=y$, namely $\frac{x}{2}+\frac{5}{2}=x^2-4x+7$
Solve for real $x$, should they exist...
As for the tangent, take the derivative of the curve with respect to $x$, namely $$y'(B)=2(B)-4$$
Evaluate $$\lim_{h->0}\frac{(B+h)^2-4(B+h)+7-B^2+4B-7}{h}$$
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for the tangent, what do you do once you get the derivative...i am still a bit confused yet... – Roslyn Jun 15 '14 at 23:29
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you use the x that you found previously? – Roslyn Jun 15 '14 at 23:44
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Yes, namely x = B. – Jun 15 '14 at 23:48
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and h is?....sorry i am not very good at understanding math – Roslyn Jun 15 '14 at 23:53
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actually i figured it out – Roslyn Jun 16 '14 at 00:25