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liberstina [14]
3 years ago
12

Graph the function f(x)= -(x-2)^2 +4. In what form is this quadratic function written? _________________________. List the value

s of the following parameters:a = ______h = ______k = ______Vertex = ___________
Mathematics
2 answers:
nevsk [136]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

f(x)= -(x-2)^2 +4

This is the vertex form of the equation for a parabola

y= a ( x-h)^2 +k  where (x,k) is the vertex of the parabola

a = -1   h= 2   k = 4

The vertex is ( 2,4)

enyata [817]3 years ago
4 0

Answer:

Completed square form

a = -1

h = 2

k = 4

Vertex: (2,4)

Step-by-step explanation:

f(x)= -(x-2)^2 +4 is in the complete square form

f(x) = a(x - h)² + k

a = -1

h = 2

k = 4

Vertex: (2,4)

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Is 7 a solution to 4x=21
kotykmax [81]

Answer:

No

Step-by-step explanation:

7 would be the solution to 3x=21, the solution to 4x=21 is 21/4 or 5.25.

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ladessa [460]

Answer:

Step-by-step explanation:

There are 6 solutions or zeros here because, according to the Fundamental Theorem of Algebra, the degree of the polynomial dictates how many zeros there are in the polynomial.  If we had a 3rd degree polynomial, we would expect to find 3 zeros; if we had a 5th degree polynomial, we would have 5 zeros, etc.  The easiest way to factor this is to do it initially by grouping:

(4x^6+20x^4)-(25x^2-125) then

4x^4(x^2+5)-25(x^2+5) then

(x^2+5)(4x^4-25)=0

We will factor each set of parenthesis now to get all the zeros.  For the first set of parenthesis:

x^2+5=0 so

x^2=-5 so

x=\sqrt{-5}

But since we can't have a negative under the square root, we have to offset it by using the imaginary number i.  i-squared = -1, so

x = ±i√5

Those are the first 2 zeros out of 6.  Now for the second set of parenthesis:

4x⁴ - 25 = 0.  That is the difference between perfect squares, and that factors to this:

(2x² + 5)(2x² - 5)

The first set of parenthesis there:

2x² + 5 = 0 so

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x = ±\sqrt{\frac{5}{2} }i

Those are the next 2 zeros.  We found 4 so far, now we will find the last 2 in the second set of parenthesis above:

x^2-5=0 so

x^2=5 so

x = ±\sqrt{\frac{5}{2} }

In summary, the 6 zeros are as follows:

x = \sqrt{5}i, -\sqrt{5}i, \sqrt{\frac{5}{2} }i, -\sqrt{\frac{5}{2} }i, \sqrt{\frac{5}{2} }, -\sqrt{\frac{5}{2} }

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