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I can't solve the last exercises in a worksheet of Pre-Calculus problems. It says:

Quadratic function $f(x)=ax^2+bx+c$ determines a parabola that passes through points $(0, 2)$ and $(4, 2)$, and its vertex has coordinates $(x_v, 0)$.

a) Calculate coordinate $x_v$ of parabola's vertex.

b) Calculate $a, b$ and $c$ coefficients.

How can I get parabola's equation with this information and find what is requested?

I would appreciate any help. Thanks in advance.

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    $\begingroup$ Hint: Plug the two points into the equation of the parabola. You get two equations in three unknowns (a, b and c). Then, you know that when the parabola is zero, it's derivative is also zero. Equate the two to get the third equation. Solve three equations in three unknowns. $\endgroup$ – Rohit Pandey Feb 24 at 21:09
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    $\begingroup$ One short-cut: $f(0)=2$ means $c = ?$ $\endgroup$ – J. W. Tanner Feb 24 at 21:10
  • $\begingroup$ @RohitPandey Not quite. The derivative is $0$ at the vertex, not when the "parabola is $0$". $\endgroup$ – Ethan Bolker Feb 24 at 21:14
  • $\begingroup$ @EthanBolker - Doesn't the vertex having a y-coordinate of $0$ imply that the parabola is $0$ there? $\endgroup$ – Rohit Pandey Feb 24 at 21:18
  • $\begingroup$ @J.W.Tanner Thanks, I tried that and $c$ equals 2 in that case. But still I have two unknown variables $a$ and $b$. How can I calculate them? $\endgroup$ – F. Zer Feb 24 at 21:30
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Since $f(0)=c$ and we are given $f(0)=2$, we see immediately that $c=2.$

Furthermore, the equation in vertex form is $f(x)=a(x-x_v)^2+k$,

and since we are given $f(x_v)=0$, we see that $k=0,$ i.e., $f(x)=a(x-x_v)^2$.

From $a(x-x_v)^2=ax^2+bx+2$ we see that $ax_v^2 = 2$ and $-2ax_v=b.$

Since $f(4)=f(0)=2$, $(4-x_v)^2=x_v^2$, which means $x_v=2$. Thus $a=\frac12$ and $b=-2.$

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  • $\begingroup$ Thank you. Very much apreciated. One question though: how did you concluded −2𝑎𝑥𝑣=𝑏 ? $\endgroup$ – F. Zer Feb 24 at 22:15
  • $\begingroup$ $ax^2-2ax_vx+ax_v^2=ax^2+bx+2$ for all $x$ implies $-2ax_v=b$ and $ax_v^2=2$ $\endgroup$ – J. W. Tanner Feb 24 at 22:18
  • $\begingroup$ I understand more, now. One last question: how did you reached $(4 - x_v)^2 = x_v^2$. Where did $a$ went ? $\endgroup$ – F. Zer Feb 25 at 13:05
  • $\begingroup$ We have $a(4-x_v^2)^2=ax_v^2;$ divide both sides by $a$ (which is not zero -- otherwise we would have a line, not a parabola) $\endgroup$ – J. W. Tanner Feb 25 at 13:51
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HINTS:

A graph is a collection of points where the $x$ and $y$ coordinates of these points are in a relationship. We sometimes write $y$ instead of $f(x)$ to stress this fact.

Your equation

$$ y=ax^2+bx+c $$ is this relation.

Try plugging in the coordinates of your given points, which you know lie on this curve (so they will satisfy the linking relation between the coordinate pairs)

I would definitely start with the point $(0,2)$, zeros are always good to have around.

You will get

$$ 2=a\cdot0^2+b\cdot0+c $$

Then I would try with the other two points.

Hope this helped

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  • $\begingroup$ Thanks. I am left with this equation $16a + 4b = 0$ if I plug known points. Still have missing pieces. $\endgroup$ – F. Zer Feb 24 at 21:35
  • $\begingroup$ @F.Zer I know, you need $3$ points to determine a parabola. But I think you made a mistake. I believe that it should not be $0$ on the right hand side $\endgroup$ – Vinyl_cape_jawa Feb 24 at 21:38
  • $\begingroup$ @Vinyl_coat_java Previous step was: $$ \left\{ \begin{array}{c} a(0)^2+b*0+c = 2, \\ a(4)^2 + b*4 + 2 = 2 \end{array} \right. $$ Do you think it is wrong? $\endgroup$ – F. Zer Feb 24 at 21:44
  • $\begingroup$ @F.Zer Correct! I made a mistake, sorry. So now you have $b=-4a$. Now you could take the quadratic equation and do something what is called "complete the squre". You know how to do that? $\endgroup$ – Vinyl_cape_jawa Feb 24 at 21:53
  • $\begingroup$ I know how to complete the square. What do you suggest? $\endgroup$ – F. Zer Feb 24 at 22:15

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