Zeros and Vertex

[dat person]

I need help understanding how to find the zeros and vertex just by looking at the equation "h=20t-5[imath]t^2[/imath]" Not sure if this is the right place to ask it in

 

[Harry_the_cat]

This is a quadratic function and the graph will be a parabola. I assume you know that already!

Write it as \(\displaystyle h= -5t^2+20t\).

This is in the standard form of a quadratic function \(\displaystyle y=ax^2+bx+c\) where \(\displaystyle c=0\).

The zeroes occur when \(\displaystyle h=-5t^2+20t = -5t(t-4)=0\) ie t=0 and t=4. (using simple factorisation)

The x (or t) coordinate of the vertex is halfway between the zeroes (in this case t=2). The y (or h) coordinate is found by substitution.

 

[BigBeachBanana]

Another way you can find the vertex of a quadratic function by: [math]\frac{-b}{2a} =\frac{20}{2*5} = 2[/math]

 

[lookagain]

Another way you can find the vertex of a quadratic function by: [math]\frac{-b}{2a} =\frac{20}{2*5} = 2[/math]

You did not show the values being correctly substituted, nor did you show
what variable this quantity represents.

\(\displaystyle h \ = \ -5t^2 + 20t \)

In this case,

\(\displaystyle t \ = \ \dfrac{-b}{2a} \)

\(\displaystyle t \ = \dfrac{-(20)}{2(-5)}\)

\(\displaystyle t \ = \ \dfrac{-20}{-10}\)

\(\displaystyle t \ = \ 2 \)

 

[Harry_the_cat]

If you've got to find the zeroes anyway, I think it is easier to find the half-way value.
I realise, of course, that if there are no zeroes then the -b/2a method will get you out of trouble. But if c=0, there will always be zeroes.

 

[Otis]

find the zeros and vertex just by looking at the equation

Hi person. We can't "see" the zeros and the vertex coordinates by just looking at that equation. There's thinking involved. (Sometimes, we get a quadratic polynomial in vertex form. In those cases, we can read off the vertex coordinates.)

It's possible to do the calculations in your head, if that's what you meant by "just looking".

?

[imath]\;[/imath]

 

[Otis]

Another way you can find the vertex of a quadratic function

Yes, that formula gives you the first part of the vertex -- the t-coordinate. (That formula is often one of the first things I think of, when I see "vertex" in a quadratic word problem.)

In this exercise, I'd thought about the zeros by using that formula. First, I noticed that each term contains a factor of t, so t=0 is one zero. Next, I used -b/(2a) to find the axis of symmetry (t=2). Then, I knew by symmetry that the other zero is t=4. Lastly, I evaluated h at t=2 in my head and made a mistake, ha.



[imath]\;[/imath]

 

[Steven G]

If you've got to find the zeroes anyway, I think it is easier to find the half-way value.
I realise, of course, that if there are no zeroes then the -b/2a method will get you out of trouble. But if c=0, there will always be zeroes.

zeros ?? Sloppy.

 

[Harry_the_cat]

zeros ?? Sloppy.

Ok. If c=0, and b\(\displaystyle \neq0\), there will always be zeroes.

Or are you questioning my spelling of zeroes? Depends what dictionary you use.

 

[Steven G]

Ok. If c=0, and b\(\displaystyle \neq0\), there will always be zeroes.

Now I am happy. Thank you for making my evening!