Please answer each question

a parabola has an x-intercept at 2, its axis of symmetry is the line x=4, and the y-coordinate of its vertex is 6. Determine the

nlexa [21]

Answer:

The standard equation of the parabola is:

$y=-\frac{3}{2}x^2+12x-18$

Step-by-step explanation:

An x intercept of 2 means that the point (2, 0) is in the graph of the parabola.

We can also write the general expression for the parabola in vertex form, since we can use the information on the coordinates of the vertex: (4, 6) - recall that the axis of symmetry of the parabola goes through the parabola's vertex, so the x-value of the vertex must be x=4.

$y-y_{vertex}=a\,(x-x_{vertex})^2\\y-6=a\,(x-4)^2$

Now we can find the value of the parameter "a" by using the extra information about the point (2, 0) at which the parabola intercepts the x-axis:

$y-6=a\,(x-4)^2\\0-6=a\,(2-4)^2\\-6=a\,4\\a=-\frac{6}{4} =-\frac{3}{2}$

Then the equation of the parabola becomes:

$y-6=-\frac{3}{2} \,(x-4)^2\\y-6=-\frac{3}{2} (x^2-8x+16)\\y-6=-\frac{3}{2}x^2+12x-24\\y=-\frac{3}{2}x^2+12x-18$

7 0

3 years ago

Using the given information, give the vertex form equation of each parabola.

Amanda [17]

Answer:

The equation of parabola is given by : $(x-4) = \frac{-1}{3}(y+3)^{2}$

Step-by-step explanation:

Given that vertex and focus of parabola are

Vertex: (4,-3)

Focus:( $\frac{47}{12}$ ,-3)

The general equation of parabola is given by.

$(x-h)^{2} = 4p(y-k)$ , When x-componet of focus and Vertex is same

$(x-h) = 4p(y-k)^{2}$ , When y-componet of focus and Vertex is same

where Vertex: (h,k)

and p is distance between vertex and focus

The distance between two points is given by :

L= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

For value of p:

p= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

p= $\sqrt{(4-\frac{47}{12})^{2}+((-3)-(-3))^{2}}$

p= $\sqrt{(\frac{1}{12})^{2}}$

p= $\frac{1}{12}$ and p= $\frac{-1}{12}$

Since, Focus is left side of the vertex,

p= $\frac{-1}{12}$ is required value

Replacing value in general equation of parabola,

Vertex: (h,k)=(4,-3)

p= $\frac{-1}{12}$

$(x-h) = 4p(y-k)^{2}$

$(x-4) = 4(\frac{-1}{12})(y+3)^{2}$

$(x-4) = \frac{-1}{3}(y+3)^{2}$

8 0

3 years ago

When 200 gallons of oil were removed from a tank, the volume of oil left in the tank was 3/7 of the tank's capacity. what was th

ki77a [65]

Answer:

Step-by-step explanation:

"When 200 gallons of oil were removed from a tank" algebraically looks like this:

V - 200.

"...the volume of oil left in the tank was 3/7 of the tank's capacity" algebraically looks like this:

3/7(V)

Therefore, the equation is

V - 200 = 3/7(V)

Begin by multiplying both sides by 7:

7(V - 200) = 3V and

7V - 1400 = 3V so

-1400 = -4V so

V = 350 gallons

That's if the volume of the oil in the tank was 3/7 of the tank's capacity.

For the other part of the problem, we set up the equation almost the same, except the 3/7 is a 1/2:

V - 200 = 1/2(V)

Multiplying both sides by 2 gives you

2(V - 200) = V and

2V - 400 = V so

-400 = -V so

V = 400

7 0

2 years ago

How would u write Subtract 2 from 5 times d

julsineya [31]

Answer:

5D-2

Step-by-step explanation:

5 times d = 5d. subtract 2 is after the 5d

5 0

2 years ago

H2 + 7h + 10 = 0<br> Help please

Leni [432]

Answer:

-2, -5

Step-by-step explanation:

h2 + 7h + 10 = 0

h^2 +7h + 10 = 0

h^2 +2h +5h + 10 = 0

h(h+2) + 5(h+2) = 0

h+2 = 0 or h+5 = 0

h = -2 or h = -5

4 0

3 years ago