Help! Solving for y in 2x^2 + 12x - 14

  • Thread starter jellybelly
  • Start date
In summary, solving for y in an equation means finding the value or values of y that make the equation true. This is done by finding the points where the graph of the equation intersects the y-axis. To solve for y in a quadratic equation, we can rearrange the equation into the form y = ax^2 + bx + c and use the quadratic formula or factor the equation. It is possible to have more than one solution for y in a quadratic equation due to the presence of two x-intercepts. The coefficients in a quadratic equation represent the rate of change, x-intercepts, and y-intercept of the parabola. While calculators can be used to solve for y in a quadratic equation, it is important to understand
  • #1
jellybelly
1
0
algebra help!

f(x) = 2x^2 + 12x - 14 *it's 2xSQUARED

how do you find the inverse of this equation??
x = 2y2 + 12y - 14
x + 14 = 2y^2 +12y ...How do you isolate "y" ??

I've been posting on as many sites as I can but no one has been able to help me!
any help is appreciated!
thankS!
kelsey
 
Mathematics news on Phys.org
  • #2
What kind of equation is x = 2y² + 12y - 14?
 
  • #3


Hi Kelsey, I'd be happy to help you with this problem! To find the inverse of an equation, we need to switch the roles of x and y. This means that x will become the dependent variable and y will become the independent variable.

To isolate y, we need to use algebraic manipulation to get y by itself on one side of the equation. Here are the steps to follow:

1. Start by subtracting 14 from both sides of the equation to get rid of the constant term on the right side.
x + 14 = 2y^2 + 12y - 14 - 14

2. Simplify the right side of the equation by combining like terms.
x + 14 = 2y^2 + 12y - 28

3. Next, we want to get rid of the coefficient of 2 on the y^2 term. To do this, we divide both sides of the equation by 2.
(x + 14)/2 = (2y^2 + 12y - 28)/2

4. Simplify the right side of the equation by dividing each term by 2.
(x + 14)/2 = y^2 + 6y - 14

5. Now we have a quadratic equation in terms of y. To solve for y, we need to use the quadratic formula: y = (-b ± √(b^2-4ac))/2a
In this case, a = 1, b = 6, and c = -14. Substituting these values into the formula, we get:
y = (-6 ± √(6^2-4(1)(-14)))/2(1)
y = (-6 ± √(36+56))/2
y = (-6 ± √92)/2
y = (-6 ± 9.6)/2

6. This gives us two possible values for y:
y = (-6 + 9.6)/2 = 1.8
y = (-6 - 9.6)/2 = -7.6

Therefore, the inverse of the given equation is:
f^-1(x) = 1.8 or -7.6

I hope this helps! Let me know if you have any further questions. Good luck with your algebra studies!
 

1. What does it mean to solve for y in an equation?

When solving for y in an equation, we are trying to find the value or values of y that make the equation true. In other words, we are looking for the points where the graph of the equation intersects the y-axis.

2. How do I solve for y in a quadratic equation?

To solve for y in a quadratic equation, we need to rearrange the equation into the form y = ax^2 + bx + c. Then, we can use the quadratic formula (y = (-b ± √(b^2 - 4ac)) / 2a) or factor the equation to find the values of y.

3. Can we have more than one solution for y in a quadratic equation?

Yes, it is possible to have more than one solution for y in a quadratic equation. This is because a quadratic equation can have two x-intercepts, which correspond to two different values of y.

4. What do the coefficients in a quadratic equation represent?

In the equation y = ax^2 + bx + c, the coefficient a represents the rate of change of the parabola (how steep or flat it is), the coefficient b represents the x-intercept(s), and the constant c represents the y-intercept.

5. Can I use a calculator to solve for y in a quadratic equation?

Yes, most scientific and graphing calculators have a function to solve for y in a quadratic equation. However, it is important to understand the steps involved in solving the equation by hand in order to fully comprehend the concept.

Similar threads

Replies
7
Views
3K
  • General Math
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
517
Replies
4
Views
1K
  • General Math
Replies
13
Views
2K
Replies
7
Views
1K
Replies
1
Views
1K
Replies
6
Views
2K
Replies
10
Views
3K
Back
Top