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2 years ago

Opens up or down, Vertex: (5,2), Passes through: (6,3)

1 answer:

4 0

I’m not sure if it’s a.) f(x)=(x-5)2+2 because it’s vertex is (5,2) although it doesn’t pass through (6,3). It passes through (6,4)
I think the answer is c.) none of these

Here’s the graph for it

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Select the statements that are true for the graph of y=(x+2)^2+4

likoan [24]

Alrigty

in form
$y=a(x-h)^2+k$
the vertex is (h,k)
the constant, a, deterimines the size and direction
if a>0, then the parabola opens up and the vertex is the minimum
if a<0 then the parabola opens down and the vertex is the maximum

so we are given
$y=1(x-(-2))^2+4$
1>0 so it opens up
vertex is (-2,4)
the vertex is a minimum

the vertex is (-2,4) and the graph has a minimum

3 0

2 years ago

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Points for whoever wants them sjakakakkaai

const2013 [10]

Answer:

thank you!!! hope you have a great day! :)

6 0

2 years ago

-4x + 11y = 15 x=2y Solve for y and x

Stolb23 [73]

Answer:

to find Y

-4x + 11y = 15

x - 2y = 0

equal= -3x + 9y= 15

since x=2y

-3(2y) + 9y = 15

-6y + 9y =15

3y = 15

y = 15/ 3

y = 5

to find x

using x=2y

since

y=5

,therefore

x= 2(5)

x= 10

8 0

2 years ago

Susie spent $78 in 3 hours. How much did Susie spend per hour?

mart [117]

Answer:

$26

Step-by-step explanation:

78 divided by 3

3 0

2 years ago

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What is limit of f(x) = (x^3-2x^2-9x+4)/(x^2-2x-8) as x approaches 4

sattari [20]

$\displaystyle\lim_{x\to4}\frac{x^3-2x^2-9x+4}{x^2-2x-8}$

First notice that $x^2-2x-8=(x-4)(x+2)$ . If $x-4$ is not a factor of the numerator, then there is a non-removable discontinuity at $x=4$ , and a removable discontinuity otherwise.

You have $4^3-2\times4^2-9\times4+4=0$ , which means, by the polynomial remainder theorem, that $x-4$ is indeed a linear factor of the numerator. Dividing yields a quotient of

$\dfrac{x^3-2x^2-9x+4}{x-4}=x^2+2x-1$

so the limit is

$\displaystyle\lim_{x\to4}\frac{x^2+2x-1}{x+2}=\frac{4^2+2\times4-1}{4+2}=\frac{23}6$

6 0

3 years ago