How can I find the vertex of a parabola using only $x$ intercepts.

Question

My teacher gave me this problem where I did a long jump and recorded the distance I went. He then asked us the height. My distance was 80 inches so the x-intercepts are $0,0$ and $80,0$. My question Is how can I make a quadratic Equation using only the $x$ intercepts to find height.

Answer 1

You need additional data to find the height.

For example, the parabolas $y=-x^2+80x$ and $y=-2x^2+160x$ both have the $x$-intercepts you specify, but vertices at $(40,1600)$ and $(40,3200)$, respectively.

Answer 2

Knowing the $x$-intercepts doesn't completely determine the parabola. If the specified intercepts are $a$ and $b$, then every function $f:\mathbb R\rightarrow \mathbb R$ with $f(x)=k(x-a)(x-b)$ has a graph that is a (possibly degenerate) parabola for which these values are intercepts.

But if, say, you also know another point $(u,v)$ that also lies on the parabola, the parabola is determined. That's because you can plug the coordinates into the equation for the parabola and find $k$ (set $x=u$ and set $f(x)=v$; this gives the equation $v=k(u-a)(u-b)$ to solve for $k$, right?).

Answer 3

Basically, you need to know three points on the parabola to determine it uniquely. The general form of a parabola is $ax^2 + bx + c$, which is three unknowns, which means we need three linear relations (given by three points on the curve) to determine it. Just knowing the x-intercepts does not give enough information to determine all three coefficients.

Answer 4

While the height cannot possibly be determined from just the two $x$ intercepts, and you thus need more information, there are quite a few single pieces of additional information that can be used, especially since we know the strength of gravity, $386 in/s^2$. Here are a few:

• A third point on the parabola This one's pretty obvious; you find the parabola through the three points and the apex will fall right out.

• Airtime Interestingly, this is independent of the distance jumped, the height is equal to $gt^2/8$, about 48 inches upward for a jump of 1 second, and about 12 inches for a jump of 1/2 second.

• Angle of liftoff The formula of the parabola is then $\tan\theta\cdot x \cdot (d-x)/d$, which gives a vertex height of $d\tan\theta/4$. In your case, for $\theta=45^\circ$ that's 20 inches upward; for a shallower $\theta=14^\circ$ it's 5 inches. This works out the same for angle of landing.

• Ground speed during the leap You can use this and the standard speed formula to find the time spent airborne, making this equivalent to the airtime method.

• Airspeed at liftoff This is the hardest. Given the airspeed, you can find the target angle of firing as $$\theta =\frac{\arcsin\left(\frac{gd}{v^2}\right)}{2}$$ which you can then use in the angle of liftoff method. Interestingly, you actually get two possible angles, the $\theta$ found above, and $90^\circ-\theta$. The latter gives a higher apex.