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Help.

NotSoSmart  Nov 1, 2017
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2+0 Answers

 #1
avatar+634 
+2

For the first, plug an x-value into each equation to determine which gives out the y-value.

\(-2.5(-2)^2+5(-2) \)

Simplify.

\(-2.5(4)-10\)

\(-10-10=-20\)

This is the y-value of (-2,-20), so it works. Now plug the x values from the other equations in.

-2.5(0)2+5(0)

Simplify.

-2.5(0)+0

0+0=0

0 is not the y-value of (0,-4), so you have to try another.

\(-(-2)^2+4(-2)-4=-4-8-4=-16\)

This is not the y-value of (-2,-20), so try another.

\(-2(-2)^2+4(-2)-4=-8-8-4=-20\)

This is the y-value of (-2,-20), so it works. Now plug the x values from the other equations in.

\(-2(0)^2+4(0)-4=0+0-4=-4\)

This is the y-value of (0,-4), so plug in the final point.

\(-2(4)^2+4(4)-4=-32+16-4=-16-4=-20\)

This is the y-value of (4,-20), so the equation is -2x2+4x-4.

AdamTaurus  Nov 2, 2017
 #2
avatar+634 
+1

For the second, use the quadratic formula.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

In this, a=4, b=11, and c=6.

\( {-11 \pm \sqrt{11^2-4(4)(6)} \over 2(4)}\)

Simplify.

\({-11 \pm \sqrt{121-16(6)} \over 8}\)

\({-11 \pm \sqrt{121-96} \over 8}\)

\({-11 \pm \sqrt{25} \over 8}\)

\(\frac{-11 \pm5}{8}\)

This is equivalent to \(\frac{-11+5}{8},\frac{-11-5}{8}\).

\(\frac{-6}{8},\frac{-16}{8}=\frac{-3}{4},-2\)

So your factors are (x+3/4)(x+2) (because the standard form is (x-a)(x-b)).

But there is one more thing. Just move the denominator in front of the x, because no one wants fractions (and also it wouldn't be right otherwise).

So your real factors are \((4x+3)(x+2)\).

AdamTaurus  Nov 2, 2017

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