Math, asked by kunichanpradeep4153, 9 months ago

The maximum or minimum point of the fraction x^2 +y^2+

Answers

Answered by Anonymous
1

Step-by-step explanation:

Given:

y

=

2

x

2

18

x

+

13

is a quadratic equation in standard form:

y

=

a

x

2

+

b

x

+

c

,

where:

a

=

2

,

b

=

18

, and

c

=

13

To graph a quadratic function, you need to have at least the vertex and x-intercepts. The y-intercept is helpful, also.

Axis of Symmetry: vertical line

(

x

,

±

)

that divides the parabola into two equal halves. The variable for the line is

x

=

b

2

a

.

x

=

(

18

)

2

2

x

=

18

4

x

=

9

2

axis of symmetry and

x

-value for the vertex

Vertex: the maximum or minimum point of the parabola. If

a

>

0

, the vertex is the minimum point and the parabola will open upward. If

a

<

0

, the vertex is the maximum point and the parabola will open downward.

We have the

x

-value of the vertex. To determine the

y

-value, substitute

9

2

for

x

in the equation and solve for

y

.

y

=

2

(

9

2

)

2

18

(

9

2

)

+

13

Simplify.

y

=

2

(

81

4

)

162

2

+

13

All terms must have a common denominator of

4

. Multiply fractions without a denominator of

4

by a multiplier equal to

1

that will produce an equivalent fraction with a denominator of

4

. For example,

3

3

=

1

Recall that any whole number,

n

, is understood to have a denominator of

1

. So

13

=

13

1

y

=

162

4

162

2

×

2

2

+

13

1

×

4

4

y

=

162

4

324

4

+

52

4

y

=

(

162

324

+

52

)

4

y

=

110

4

y

=

55

2

Vertex:

(

9

2

,

55

2

)

minimum point of the parabola

Approximate vertex:

(

4.5

,

27.5

)

Substitute

0

for

y

and use the quadratic formula to find the roots and the x-intercepts.

0

=

2

x

2

18

x

+

13

x

=

b

±

b

2

4

a

c

2

a

Plug in the known values.

x

=

(

18

)

±

(

18

)

2

4

2

13

2

2

Simplify.

x

=

18

±

324

104

4

x

=

18

±

220

4

Prime factorize

220

.

x

=

18

±

(

2

×

2

)

×

5

×

11

4

x

=

18

±

2

55

4

Simplify.

x

=

9

±

55

2

Roots: values for

x

x

=

9

+

55

2

,

9

55

2

Approximate values for

x

.

x

=

8.21

,

0.792

X-intercepts: values of

x

when

y

=

0

x

-intercepts:

(

9

+

55

2

,

0

)

and

(

9

55

2

,

0

)

Approximate

x

-intercepts:

(

8.21

,

0

)

and

(

0.792

,

0

)

Y-Intercept: value of

y

when

x

=

0

y

=

2

(

0

)

2

18

(

0

)

+

13

y

=

13

Y-intercept:

(

0

,

13

)

Plot the vertex and x-intercepts and sketch a parabola through the points. Do not connect the dots.

graph{y=2x^2-18x+13 [-13.95, 18.07, -40.31, -24.29]}

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