SOLUTION: for the quadratic equation -16t^2+125t what is the maximum and minimum?

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Question 618720: for the quadratic equation -16t^2+125t what is the maximum and minimum?
Found 3 solutions by Alan3354, josmiceli, math-vortex:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
for the quadratic equation -16t^2+125t what is the maximum and minimum?
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No equation is given.
-16t^2+125t is a binomial, not an equation.
Equations have equal signs.
-16t^2+125t might = 0
-16t^2+125t might = -1000
etc

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The minus in front of the x-squared term tells me
there is a maximum and not a minimum
The maximum is halfway between the 2 roots
+-16t%5E2+%2B+125t+=+0+
+t%2A%28+-16t+%2B+125+%29+=+0+
+-16t+=+-125+
+t+=+125%2F16+
+t+=+7.8125+
One root is at +t+=+0+ also, so the
halfway point is +%28+7.8125+%2B+0+%29+%2F+2+=+3.90625+
This is the value of +t+
The value of the function is:
+f%28t%29+=+-16%2A%283.90625%29%5E2+%2B+125%2A3.90625+
+f%28t%29+=+-+244.141+%2B+488.282+
+f%28t%29+=+244.141+
The maximum is at ( 3.907, 244.141 ) approximately

Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--
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A quadratic equation only has one, either a maximum or a minimum. It won't have both.
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The shape of the graph of a quadratic is a parabola. If the parabola opens upward, the the equation has a minimum at the vertex. If the parabola opens downward, the equation has a maximum at the vertex.
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There are many ways to solve this problem. I'm not sure what math level you are studying, so I'll show an Algebra I method.
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Since the minimum or maximum is located at the vertex, we'll end the vertex of this equation. The vertex will be an ordered pair (t,h). We have
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h=-16t%5E2%2B125t
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Notice that I added an h to be the 2nd variable of the equation. You could use any variable. When a quadratic equation is in standard form,
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h=at%5E2%2Bbt%2Bc
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the t-value of the vertex is the value -b/2a, where a and b are the coefficients of the t^2 term and the t term in your equation. In your case a=-16 and b=125. Therefore,
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-b%2F2a=%28-125%29%2F%282%2A-16%29=125%2F32
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So the t-value of the vertex is 125/32. To find the h-value, we use substitution:
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h=-16%28125%2F32%29%5E2%2B125%28125%2F32%29
h=-250000%2F1024%2B15625%2F32
h=-250000%2F1024%2B500000%2F1024
h=250000%2F1024=15625%2F64
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Now we know that the vertex of the parabola is at the point (125/32, 15625/64).
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When a quadratic equation is in standard form, we can use the leading coefficient---the a-value---to determine if the parabola opens upward or downward. If a>0, the parabola opens up; if a<0, it opens down. In your case the parabola opens downward. This means that the vertex is a maximum for your parabola.
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Hope this helps. Feel free to email if you have questions about this.
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Ms.Figgy
math.in.the.vortex@gmail.com