Art, asked by hlloyd7671, 1 month ago

Which equation is y = (x + 3)2 + (x + 4)2 rewritten in vertex form?

Answers

Answered by dk1789774
6

Answer:

we have

y=(x+3)^{2}+(x+4)^{2}y=(x+3)

2

+(x+4)

2

\begin{gathered}y=(x+3)^{2}+(x+4)^{2} \\y= x^{2} +6x+9+x^{2}+8x+16 \\y=2x^{2}+14x+25\end{gathered}

y=(x+3)

2

+(x+4)

2

y=x

2

+6x+9+x

2

+8x+16

y=2x

2

+14x+25

rewrite now in vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

y-25=2x^{2}+14xy−25=2x

2

+14x

Factor the leading coefficient

y-25=2(x^{2}+7x)y−25=2(x

2

+7x)

Complete the square. Remember to balance the equation by adding the same constants to each side

y-25+24.50=2(x^{2}+7x+3.5^{2})y−25+24.50=2(x

2

+7x+3.5

2

)

y-0.50=2(x^{2}+7x+12.25)y−0.50=2(x

2

+7x+12.25)

Rewrite as perfect squares

y-0.50=2(x+3.5)^{2}y−0.50=2(x+3.5)

2

y=2(x+3.5)^{2}+0.50y=2(x+3.5)

2

+0.50

the vertex is the point (-3.5,0.50)(−3.5,0.50)

therefore

the answer is

y=2(x+3.5)^{2}+0.50y=2(x+3.5)

2

+0.50

Answered by mahighagargunde
0

Answer:

This is a bit of a sneaky question. It isn't immediately obvious that this is a parabola, but "vertex form" is a form of equation specifically for one. It is a parabola, a closer look reveals, which is fortunate... It's the same thing as "completing the square" - we want the equation in the form

a

(

x

h

)

2

+

k

.

To get there from here, we first multiply out the two brackets, then collect up terms, then divide through to make the

x

2

coefficient 1:

1

2

y

=

x

2

+

7

x

+

25

2

Then we find a square bracket that gives us the correct

x

coefficient. Note that in general

(

x

+

n

)

2

=

x

2

+

2

n

+

n

2

So we choose

n

to be half our existing

x

coefficient, i.e.

7

2

. Then we need to subtract off the extra

n

2

=

49

4

that we've introduced. So

1

2

y

=

(

x

+

7

2

)

2

49

4

+

25

2

=

(

x

+

7

2

)

2

+

1

4

Multiply back through to get

y

:

y

=

2

(

x

+

7

2

)

2

+

1

2

Answer link

Similar questions