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3 years ago

How would you determine which x-values to use when creating a table of values to graph a parabola?

1 answer:

5 0

To determine which values of x we would use for creating a graph of a parabola, we need to know where the line of symmetry, or the axis of symmetry is. For that we can use the equation:

y=(x-h)+k, where we know h and k.

From this equation we can see that the line of symmetry is passing trough x=h.

And now we can determine which values do we need to add to h and to subtract from h to get values of x to create the table of values to plot a parabola.

You might be interested in

Each triangle weighs 8 grams. Each circle weighs 3 grams. Each trapezoid weighs x grams.

sesenic [268]

Answer:

Trapezoid = 24 grams

Step-by-step explanation:

Trapezoid = 3 triangles

x = 3 · 8

x = 24

4 0

3 years ago

Solve each linear equation for the variable.

Nikitich [7]

X + 2 (2x + 20) = 85
x + 4x + 40 = 85
5x + 40 = 85
5x = 45
x = 9

12y + 6 + 32y = 2y - 6y - 30
44y + 6 = -4y -30
48y = -36
y = -36/48
y = -3/4

7 0

4 years ago

Read 2 more answers

An airplane is headed on a bearing of 159° at an airspeed of 211 km per hr. A 44 km per hr wind is blowing from a direction of 2

oee [108]

Answer:

The ground speed is 189Kmph

The bearing 162.37 degrees

Step-by-step explanation:

Find attached the step by step explanation with diagrams. Please respond if you need more clarification.

6 0

3 years ago

SHOW YOUR WORK!!

Dmitrij [34]

Answer:

x = -0.846647 or x = -0.177346 or x = 0.841952 or x = 1.58204

Step-by-step explanation:

Solve for x:

5 x^4 - 7 x^3 - 5 x^2 + 5 x + 1 = 0

Eliminate the cubic term by substituting y = x - 7/20:

1 + 5 (y + 7/20) - 5 (y + 7/20)^2 - 7 (y + 7/20)^3 + 5 (y + 7/20)^4 = 0

Expand out terms of the left hand side:

5 y^4 - (347 y^2)/40 - (43 y)/200 + 61197/32000 = 0

Divide both sides by 5:

y^4 - (347 y^2)/200 - (43 y)/1000 + 61197/160000 = 0

Add (sqrt(61197) y^2)/200 + (347 y^2)/200 + (43 y)/1000 to both sides:

y^4 + (sqrt(61197) y^2)/200 + 61197/160000 = (sqrt(61197) y^2)/200 + (347 y^2)/200 + (43 y)/1000

y^4 + (sqrt(61197) y^2)/200 + 61197/160000 = (y^2 + sqrt(61197)/400)^2:

(y^2 + sqrt(61197)/400)^2 = (sqrt(61197) y^2)/200 + (347 y^2)/200 + (43 y)/1000

Add 2 (y^2 + sqrt(61197)/400) λ + λ^2 to both sides:

(y^2 + sqrt(61197)/400)^2 + 2 λ (y^2 + sqrt(61197)/400) + λ^2 = (43 y)/1000 + (sqrt(61197) y^2)/200 + (347 y^2)/200 + 2 λ (y^2 + sqrt(61197)/400) + λ^2

(y^2 + sqrt(61197)/400)^2 + 2 λ (y^2 + sqrt(61197)/400) + λ^2 = (y^2 + sqrt(61197)/400 + λ)^2:

(y^2 + sqrt(61197)/400 + λ)^2 = (43 y)/1000 + (sqrt(61197) y^2)/200 + (347 y^2)/200 + 2 λ (y^2 + sqrt(61197)/400) + λ^2

(43 y)/1000 + (sqrt(61197) y^2)/200 + (347 y^2)/200 + 2 λ (y^2 + sqrt(61197)/400) + λ^2 = (2 λ + 347/200 + sqrt(61197)/200) y^2 + (43 y)/1000 + (sqrt(61197) λ)/200 + λ^2:

(y^2 + sqrt(61197)/400 + λ)^2 = y^2 (2 λ + 347/200 + sqrt(61197)/200) + (43 y)/1000 + (sqrt(61197) λ)/200 + λ^2

Complete the square on the right hand side:

(y^2 + sqrt(61197)/400 + λ)^2 = (y sqrt(2 λ + 347/200 + sqrt(61197)/200) + 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200)))^2 + (4 (2 λ + 347/200 + sqrt(61197)/200) (λ^2 + (sqrt(61197) λ)/200) - 1849/1000000)/(4 (2 λ + 347/200 + sqrt(61197)/200))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 347/200 + sqrt(61197)/200) (λ^2 + (sqrt(61197) λ)/200) - 1849/1000000 = (8000000 λ^3 + 60000 sqrt(61197) λ^2 + 6940000 λ^2 + 34700 sqrt(61197) λ + 6119700 λ - 1849)/1000000 = 0.

Thus the root λ = (-3 sqrt(61197) - 347)/1200 + 1/60 (-i sqrt(3) + 1) ((3 i sqrt(622119) - 4673)/2)^(1/3) + (19 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(622119) - 4673)^(1/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + sqrt(61197)/400 + λ)^2 = (y sqrt(2 λ + 347/200 + sqrt(61197)/200) + 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200)))^2

Take the square root of both sides:

y^2 + sqrt(61197)/400 + λ = y sqrt(2 λ + 347/200 + sqrt(61197)/200) + 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200)) or y^2 + sqrt(61197)/400 + λ = -y sqrt(2 λ + 347/200 + sqrt(61197)/200) - 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200))

Solve using the quadratic formula:

y = 1/40 (sqrt(2) sqrt(400 λ + 347 + sqrt(61197)) + sqrt(2) sqrt(347 - sqrt(61197) - 400 λ + 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197)))) or y = 1/40 (sqrt(2) sqrt(400 λ + 347 + sqrt(61197)) - sqrt(2) sqrt(347 - sqrt(61197) - 400 λ + 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197)))) or y = 1/40 (sqrt(2) sqrt(347 - sqrt(61197) - 400 λ - 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197))) - sqrt(2) sqrt(400 λ + 347 + sqrt(61197))) or y = 1/40 (-sqrt(2) sqrt(400 λ + 347 + sqrt(61197)) - sqrt(2) sqrt(347 - sqrt(61197) - 400 λ - 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197)))) where λ = (-3 sqrt(61197) - 347)/1200 + 1/60 (-i sqrt(3) + 1) ((3 i sqrt(622119) - 4673)/2)^(1/3) + (19 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(622119) - 4673)^(1/3))

Substitute λ = (-3 sqrt(61197) - 347)/1200 + 1/60 (-i sqrt(3) + 1) ((3 i sqrt(622119) - 4673)/2)^(1/3) + (19 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(622119) - 4673)^(1/3)) and approximate:

y = -1.19665 or y = -0.527346 or y = 0.491952 or y = 1.23204

Substitute back for y = x - 7/20:

x - 7/20 = -1.19665 or y = -0.527346 or y = 0.491952 or y = 1.23204

Add 7/20 to both sides:

x = -0.846647 or y = -0.527346 or y = 0.491952 or y = 1.23204

Substitute back for y = x - 7/20:

x = -0.846647 or x - 7/20 = -0.527346 or y = 0.491952 or y = 1.23204

Add 7/20 to both sides:

x = -0.846647 or x = -0.177346 or y = 0.491952 or y = 1.23204

Substitute back for y = x - 7/20:

x = -0.846647 or x = -0.177346 or x - 7/20 = 0.491952 or y = 1.23204

Add 7/20 to both sides:

x = -0.846647 or x = -0.177346 or x = 0.841952 or y = 1.23204

Substitute back for y = x - 7/20:

x = -0.846647 or x = -0.177346 or x = 0.841952 or x - 7/20 = 1.23204

Add 7/20 to both sides:

Answer: x = -0.846647 or x = -0.177346 or x = 0.841952 or x = 1.58204

5 0

4 years ago

For all the functions of the form f(x) = ac^2 + bx + c, which is true when b=0

bixtya [17]

You didn't show us the choices.
The graph of that function would be centered on the y-axis.

4 0

3 years ago

Read 2 more answers