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3 years ago

What is going f(x) = 2x2+28x-5 written is vertex form?

Mathematics

1 answer:

AVprozaik [17]3 years ago

8 0

f(x) = 2(x^2 + 14x + ) - 5 2 is taken out of first 2 terms.

f(x) = 2(x^2 + 14x + (14/2)^2 ) - 5 - 2(14/2)^2

<em><u>2 steps </u></em>

Inside the brackets: Linear term's coefficient (14) divided by 2 and squared
Outside the brackets: (14/2)^2 multiplied by 2 and subtract so that there is no change in the value of the quadratic.

f(x) = 2*(x + 7)^2 - 103

The graph is of the original quadratic.

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write the equation in slope intercept form of the line that has a slope of 2 and contains the point (1,1)

Genrish500 [490]

y = 2x + b

1 = 2x 1 + b

1 = 2 + b

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3 years ago

The equation of a parabola is given.

BlackZzzverrR [31]

Answer:

Directrix equation: y = 11/2

y = k - c = 6 - 1/2 = 11/2

Step-by-step explanation:

y=(1/2) x^2+6x+24

factor this

y = (1/2)* [ x^2 + 12x ] + 24

y = (1/2)* [ x^2 + 12x + 36 - 36] + 24

y = (1/2)* [ (x + 6)^2 - 36] + 24

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4 years ago

What is the answer to the following?

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Answer:

Step-by-step explanation:

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3 years ago

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3 years ago

Evaluate 1^3 + 2^3 +3^3 +.......+ n^3

Molodets [167]

Notice that

$(n+1)^4-n^4=4n^3+6n^2+4n+1$

so that

$\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

We have

$\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)$

$\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1$

so that

$\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

You might already know that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

so from these formulas we get

$\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n$

$\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4$

$\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}$

If you don't know the formulas mentioned above:

The first one should be obvious; if you add $n$ copies of 1 together, you end up with $n$ .
The second one is easily derived: If $S=1+2+3+\cdots+n$ , then $S=n+(n-1)+(n-2)+\cdots+1$ , so that $2S=n(n+1)$ or $S=\dfrac{n(n+1)}2$ .
The third can be derived using a similar strategy to the one used here. Consider the expression $(n+1)^3-n^3=3n^2+3n+1$ , and so on.

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4 years ago