Solve for y in the two equations below using substitution.

Write a quadratic function that only has one root

AlekseyPX

Answer:

y= x squared.

Step-by-step explanation:

When graphed, the parabola only has an x-intercept at (0,0). Since a root is another word for an x-intercept, it only has one root.

I hope this helps, and I hope you have a good day.

3 0

3 years ago

Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”

Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± $\sqrt{1-(y-2)^{2}}$

y = ± $\sqrt{1-x^{2}}+2$

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

(x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± $\sqrt{1-(y-2)^{2}}$

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± $\sqrt{1-x^{2}}$

- Add 2 for both sides

∴ y = ± $\sqrt{1-x^{2}}+2$

∴ y = h(x)

- In the function x = ± $\sqrt{1-(y-2)^{2}}$

∵ $\sqrt{1-(y-2)^{2}}$ ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± $\sqrt{1-x^{2}}+2$

∵ $\sqrt{1-x^{2}}$ ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0

4 years ago

What is the image point of (2,4) after a translation left 4 units and down 1 unit? hat is the image point of (0,1) after a trans

denis-greek [22]

Want me to help

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

You have 1 case of soap bars and there are 150 bars in a case. You use 300 bars of soap per day. How many cases do you need to o

Grace [21]

multiply 300 by 7 and then you should be able to finish the equation

5 0

3 years ago

Read 2 more answers

I WILL GIVE YOU BRAINLIEST NEED HELP PLEASEE :DD

Shtirlitz [24]

The coordinates of the pre-image of point F' is (-2, 4)

<h3>How to determine the coordinates of the pre-image of point F'?</h3>

On the given graph, the location of point F' is given as:

F' = (4, -2)

The rule of reflection is given as

Reflection across line y = x

Mathematically, this is represented as

(x, y) = (y, x)

So, we have

F = (-2, 4)

Hence, the coordinates of the pre-image of point F' is (-2, 4)