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mariarad [96]
3 years ago
9

Re-write the quadratic function below in Standard Form y = −2(x−6)(x−2)

Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

THE answer is y= -2x²+16x-24

Step-by-step explanation:

OK, so the standard form for an equation is ax²+bx+c

When you look at the equation we see that there is a -2 all by itself so we need too bring it within the parenthesis, so we'll distribute it..... so now instead of 2(x-6), we have -2x-12..........Why? because (2 times x is 2x and 2 times six is 12)

next when you write it ALL out, you get (-2x+12)(x-2) now we need to put this in the standard form by using the FOIL method  

F- first O-outer I-inner L-last..... so multiplying the First from both equations we get -2x², Outer is -2x times -2 which is 4 because two negatives make a positive, Inner is 12 times x which is 12x, Last is 12 times -2 whchis -24. Put them all together and you get. 2x²+12x+4x- 24, now we combine like terms (in bold) and we get -2x²+16x-24. Look at the standard form i gave above and see how the final answer matches

-2x²+16x-24

ax² +bx   + c

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HACTEHA [7]

Answer:

The given function f(x)=\frac{2x}{3}-4 was replaced with f(x+k)=\frac{2x}{3}-6 then the value of k is -3

that is k=-3.

Step-by-step explanation:

The given function is f(x)=\frac{2x}{3}-4\hfill (1)

and when x=x+k then f(x+k)=\frac{2x}{3}-6\hfill (2)

Now to find the value of k:

Put x=x+k in the function f(x)

f(x)=\frac{2x}{3}-4

f(x+k)=\frac{2(x+k)}{3}-4

f(x+k)=\frac{2x+2k}{3}-4\hfill (3)

Now comparing equations (2) and (3) we get

f(x+k)=\frac{2x+2k}{3}-4=\frac{2x}{3}-6

\frac{2x+2k}{3}-4=\frac{2x}{3}-6

\frac{2x+2k}{3}-4-\frac{2x}{3}+6=0

\frac{2x}{3}+\frac{2k}{3}-4-\frac{2x}{3}+6=0

\frac{2k}{3}+2=0

\frac{2k}{3}=-2

k=\frac{-2\times 3}{2}

k=-3

Therefore k=-3

Therefore the given function f(x)=\frac{2x}{3}-4 was replaced with  f(x+k)=\frac{2x}{3}-6 then the value of k is -3

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The given statement is true.

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Quadrilaterals are similar if their corresponding sides are proportional.

This statement is true.

Quadrilaterals are similar when

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So, the given statement is true.

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