# Equation of parabola that passes through two points and vertex has coordinates ($x_v$, $0$)

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.

• 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. – Rohit Pandey Feb 24 at 21:09
• One short-cut: $f(0)=2$ means $c = ?$ – J. W. Tanner Feb 24 at 21:10
• @RohitPandey Not quite. The derivative is $0$ at the vertex, not when the "parabola is $0$". – Ethan Bolker Feb 24 at 21:14
• @EthanBolker - Doesn't the vertex having a y-coordinate of $0$ imply that the parabola is $0$ there? – Rohit Pandey Feb 24 at 21:18
• @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? – F. Zer Feb 24 at 21:30

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 $$b=-2.$$

• Thank you. Very much apreciated. One question though: how did you concluded −2𝑎𝑥𝑣=𝑏 ? – F. Zer Feb 24 at 22:15
• $ax^2-2ax_vx+ax_v^2=ax^2+bx+2$ for all $x$ implies $-2ax_v=b$ and $ax_v^2=2$ – J. W. Tanner Feb 24 at 22:18
• I understand more, now. One last question: how did you reached $(4 - x_v)^2 = x_v^2$. Where did $a$ went ? – F. Zer Feb 25 at 13:05
• 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) – J. W. Tanner Feb 25 at 13:51

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.

$$y=ax^2+bx+c$$ 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

• Thanks. I am left with this equation $16a + 4b = 0$ if I plug known points. Still have missing pieces. – F. Zer Feb 24 at 21:35
• @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 – Vinyl_cape_jawa Feb 24 at 21:38
• @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? – F. Zer Feb 24 at 21:44
• @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? – Vinyl_cape_jawa Feb 24 at 21:53
• I know how to complete the square. What do you suggest? – F. Zer Feb 24 at 22:15