Question 618720
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^2+125t}}}
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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^2+bt+c}}}
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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/2a=(-125)/(2*-16)=125/32}}}
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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(125/32)^2+125(125/32)}}}
{{{h=-250000/1024+15625/32}}}
{{{h=-250000/1024+500000/1024}}}
{{{h=250000/1024=15625/64}}}
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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