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lord [1]
3 years ago
14

What is the max or min and range of y=2x^2+20x-12

Mathematics
1 answer:
shtirl [24]3 years ago
4 0

Answer: its x = 3

Step-by-step explanation:

The maximum or minimum of a quadratic function occurs at

x

=

−

b

2

a

. If

a

is negative, the maximum value of the function is

f

(

−

b

2

a

)

. If

a

is positive, the minimum value of the function is

f

(

−

b

2

a

)

.

f

min

x

=

a

x

2

+

b

x

+

c

occurs at

x

=

−

b

2

a

Find the value of

x

equal to

−

b

2

a

.

x

=

−

b

2

a

Substitute in the values of

a

and

b

.

x

=

−

−

12

2

(

2

)

Remove parentheses.

x

=

−

−

12

2

(

2

)

Simplify

−

−

12

2

(

2

)

.

Tap for more steps...

x

=

3

The maximum or minimum of a quadratic function occurs at

x

=

−

b

2

a

. If

a

is negative, the maximum value of the function is

f

(

−

b

2

a

)

. If

a

is positive, the minimum value of the function is

f

(

−

b

2

a

)

.

f

min

x

=

a

x

2

+

b

x

+

c

occurs at

x

=

−

b

2

a

Find the value of

x

equal to

−

b

2

a

.

x

=

−

b

2

a

Substitute in the values of

a

and

b

.

x

=

−

−

12

2

(

2

)

Remove parentheses.

x

=

−

−

12

2

(

2

)

Simplify

−

−

12

2

(

2

)

.

Tap for more steps...

x

=

3

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Use synthetic division and the Remainder Theorem to find P(a). P(x) = 2x3 + 4x2 − 10x − 9; a = 3
guapka [62]

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8 0
3 years ago
Chelsea drew two parallel lines KL and MN intersected by a transversal PQ, as shown below:
Ilia_Sergeevich [38]

Answer:

The correct answer choice would be:

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Hope this helps! :)

5 0
3 years ago
Read 2 more answers
) Find the coordinates of the point A' which is the symmetric point
soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point  A(3, 2) with respect to the line 2x + y - 12 = 0.​

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

C( \dfrac{x+3}{2}, \dfrac{y+2}{2})

Also, product of slope will be -1 .( Since, they are parallel )

\dfrac{y-2}{x-3} \times -2  = -1\\\\2y - 4 = x - 3\\\\2y - x = 1

x = 2y - 1

So, C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})

Also, C satisfy given line :

2\times ( \dfrac{2y + 2}{2})  + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}

Also,

x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}

Therefore, the symmetric points isA'(\dfrac{31}{5}, \dfrac{18}{5}) .

7 0
3 years ago
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