Can someone help me write this equation in standard form: x= y^2 + 14y + 20. I'm having trouble with this warm up. I know the answer is × = (y +7)^2 - 29. I'm just not sure how to get to that answer. Here's what I got so far......
x= y^2 + 14y + 20
x= y^2 + 7y + 10
There are three ways I can interpret what you are saying. It can mean that you want vertex form (which is the way the answer is written) with x isolated, or your given answer is incorrect and you want the standard from (y=ax^2 + bx + c) with y isolated, or you want y - isolated in vertex form.
Here is the solution for vertex form with x-isolated:
So, when you see a problem with a binomial squared, you should focus on that area first. How can we make a binomial with y^2 and 14y? (I am ignoring the constant, because we can add and subtract constants to make our ends meet, but we can't do the same with variables). Because the coefficient of the y is one, the general format is (y+c)(y+c)=y^2+2cy+c^2 (this is the binomial expansion.) So, we see that the constant needed for the square binomial is c^2 and the coefficient of y is 2c. So, 2c=14 and c=7 and c^2 = 49. This is true for all binomial squared, the coefficient of the x is two times the 'c.' This is apparent from the expansion we did earlier.
So now, we have to make y^2+14y+49 from the given form.
I think you got it from here, but here is a hint: 20=49-29
For y-isolated standard form:
Use the quadratic formula to solve for y in terms of x, and put in the form of ax^n + bx^n-1 + cx^n-2. Basically, order the variables from greatest to smallest order with the constants at the end.
For the y-isolated vertex form:
Use the standard form and convert that to y= (x-c)^n + k, where n is any order.
However, I am guessing vertex form with x isolated to be the most probable answer.
x= y^2 + 14y + 20
To complete the square you need to halve the coefficient of x and add the square of that. In this case, 14/2 = 7, and 7^2 = 14, so to leave things unchanged, you have to add and subtract the 49.
x = (y^2 + 14y + 49) + 20 - 49
x = (y+7)^2 = 29
I appreciate your help very much, but x= (y +7)^2 - 29 is the answer given at the back of my textbook for this equation. In the example problems it says that I should use complete the square. The only problem is that all the examples have problems like: 2x - y^2 = 2y + 7 or y = 3x^2 + 12x +9.
I've tried everything to figure out how to get to the answer my textbook gives, but I'm so confused.
To write the equation x = y^2 + 14y + 20 in standard form, which is in the form of (y - k)^2 = 4p(x - h), you need to complete the square.
Here are the steps to follow:
Step 1: Group the terms involving y.
x = (y^2 + 14y) + 20
Step 2: Take half of the coefficient of y, square it, and add it both inside and outside the parentheses.
x = (y^2 + 14y + 49 - 49) + 20
Step 3: Rearrange the terms inside the parentheses to complete the square.
x = (y^2 + 14y + 49) - 49 + 20
Step 4: Simplify the equation inside the parentheses.
x = (y + 7)^2 - 29
So, the equation x = y^2 + 14y + 20 can be rewritten in standard form as x = (y + 7)^2 - 29.