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find the equation of parabola with given two points B (2, 1) and C (4, 3) and slope of the tangent line to the parabola matches the slope of the line goes through A (0, 1.5) and B (2, 1).

i have calculated, that the slope for the line is -1/4. is it correct?
but i have no idea what the next step should be.
Any help would be appreciated.

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  • $\begingroup$ Slope of the tangent line at which point? $\endgroup$ – Hirshy Jun 14 '15 at 19:56
  • $\begingroup$ It seems like too much of a less information, firstly your parabola can be $(y-h)^2=4a(x-k)$ or it can be $(x-h)^2=4a(y-k)$ $\endgroup$ – Mann Jun 14 '15 at 19:56
  • $\begingroup$ The slope you calculated is correct $\endgroup$ – Peter Jun 14 '15 at 19:59
  • $\begingroup$ there is nothing more said. just that it is the slope of a tangent line to the parabola. $\endgroup$ – Sarah Jun 14 '15 at 19:59
  • $\begingroup$ so the derivative of the parabola equation (y=ax^2+bx+c) should be equal to -1/4? 2ax+b=-1/4 ? and what will be the x here? $\endgroup$ – Sarah Jun 14 '15 at 20:02
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To answer this problem, you must derive a system of linear equations to solve for the three unknowns in your parabola $f(x),$ where $$f(x)=ax^2+bx+c. $$ We know $f(2)=1=4a+2b+c$, $f(4)=3=16a+4b+c$, and $f'(\text{B})=-\frac{1}{4}=4a+b$. Thus, the system $$\begin{cases}4a+2b+c=1\\ 16a+4b+c=3\\ 4a+b=-\frac{1}{4} \end{cases} $$ is established. I will leave it up to you to solve for $f(x).$

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$$y=ax^2+bx+c$$ use the two points to get $$1=4a+2b+c$$ $$3=16a+4b+c$$

I think that the tangent point at point $B$ $$y'=2ax+b$$

$$-1/4=2*a*2+b$$

solve the simultaneous equation to get $a$, $b$, and $c$

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