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3 years ago

The vertex of this parabola is at (-1,-3) which of the following could be its equation

2 answers:

6 0

Since these are parabolas with y being squared, the standard form of such a parabola with vertex at (h, k) is

x = a(y - k)^2 + h 

There are no steps. Just look at what is inside the parentheses and compare it to (y - k) with k = -3. 
Then look at what is added and compare it to h = -1. 

For instance, A has (y + 1) in parentheses. So k = -1. And it has -3 added, so h = -3. That would be a vertex at (-3, -1).

inn [45]3 years ago

3 0

For
y=a(x-h)^2+k
vertex is (h,k)

it could be
y=a(x+1)^2-3

You might be interested in

Evaluate f(x) = x2 – 12 when x = 5

Ede4ka [16]

Answer:

f(5) = 13

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Function notation and substitution

Step-by-step explanation:

Step 1: Define

f(x) = x² - 12

x = 5

Step 2: Evaluate

Substitute: f(5) = 5² - 12
Exponents: f(5) = 25 - 12
Subtract: f(5) = 13

4 0

3 years ago

Read 2 more answers

How would you go about finding the solution to this system of three equations in three variables? Be specific. For example, you

kupik [55]

I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6

you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5

II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7

then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15

3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1

insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4

insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2

so the solution is
x=2
y=-4
z=-1

3 0

3 years ago

Can you help meee get this ?

nikdorinn [45]

Answer: y=4/5x-2 or y=4/5x+(-2)

Step-by-step explanation:

The formula for slope-intercept form is y=mx+b.

To find the slope, we can use the formula $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$ and plugging in the points given.

$m=\frac{6-2}{10-5} =\frac{4}{5}$

We know our slope is 4/5. We can plug this into our slope-intercept form and then plug in a point to find b, y-intercept.

$y=\frac{4}{5} x+b$

$2=\frac{4}{5}(5)+b$

$2=4+b$

$b=-2$

We know the y-intercept is -2.

THe final equation is y=4/5x-2 or y=4/5x+(-2).

4 0

4 years ago

EF is a median of trapezoid ABCD. what is the value of x?

goldfiish [28.3K]

Since the given figure is a trapezoid, here is how we are going to find for the value of x. Firstly, the sum of the bases of the trapezoid is always equal to twice of the median. So it would look like this. 2M = A + B.
Plug in the given values above.
2M = (3x+1) + (7x+1)
2(10) = 10x + 2
20 = 10x + 2
20 - 2 = 10x
18 = 10x < divide both sides by 10 and we get,
1.8 = x
Therefore, the value of x in the given trapezoid is 1.8. Hope this is the answer that you are looking for.

8 0

3 years ago

Write an equation of the line that passes through the points (0, 3) and (-5, 0).

Setler [38]

Answer:

y-3=3/5x

Step-by-step explanation:

y-y1=m(x-x1)

Where m=slope and (x1, y1) is a point on the line.

m=(y2-y1)/(x2-x1)

m=(0-3)/(-5-0)

m=-3/-5

m=3/5

y-3=3/5(x-0)

y-3=3/5(x)

y-3=3/5x

5 0

3 years ago

Read 2 more answers