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Sonbull [250]
3 years ago
11

After transforming f(x) = 2x² +4x + 3 into vertex form, the vertex is easily identifiable. Which ordered pair is the vertex?

Mathematics
1 answer:
jolli1 [7]3 years ago
7 0

The vertex is (-1,1)

Explanation:

The equation is f(x)=2x^{2} +4x+3

To find the vertex, we need the equation in the form of f(x)=a (x-h)^{2}+k

Dividing each term by 2 in the equation f(x)=2x^{2} +4x+3

f(x)=2(x^{2} +2x+\frac{3}{2} )

Now, completing the square by adding and subtracting 1, we get,

f(x)=2(x^{2} +2x+1-1+\frac{3}{2} )

The first three terms can be written as (x+1)^{2},

f(x)=2[(x+1)^{2}+\frac{1}{2}  ]

Multiplying 2 into the bracket, we get,

f(x)=2(x+1)^{2} +1

This equation is of the form f(x)=a (x-h)^{2}+k

Now, we shall find the vertex (h,k)

Thus, h=-1 and k=1

Thus, the vertex is (-1,1)

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Step-by-step explanation:

First, we have to make both the denominators the same. In this case, we can multiply 3/4 by 2, to make 4, the denominator, into 8. What we do to the denominator, we must do to the numerator! So, we multiply 3 by 2, and get 6!

Now we have, 5/8 + 6/8, which equals 11/8! Which can also be simplified to 1 and 3/8 (because the numerator is larger than the denominator)

Hope this helps! :)

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Let's assume a point (x,y) on parabola.

According to definition of parabola, the distance between point (x,y) and focus (-5,5) would be same as the distance between the point (x,y) and directrix y = -1.

\sqrt{(x+5)^2+(y-5)^2} = \sqrt{(y+1)^2} \\\\(x+5)^2+(y-5)^2=(y+1)^2 \\\\(x+5)^2 + y^2 - 10y +25 = y^2 +2y +1 \\\\(x+5)^2 =  -y^2 +10y -25 + y^2 +2y +1 \\\\(x+5)^2 = 12y -24 \\\\12y =  (x+5)^2  +24 \\\\ y = \frac{1}{12} (x+5)^2  +2

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