Write in standard form:
-2y^2+x-4y+1=0
Standard form is usually Ax + By + C = 0 for a linear function.
For a quadratic it is usually Ay^2 + By + c = 0
you have a second variable. How does your textbook deal with this?
well this is supposed to be a parabola, and i just need help with factoring it into an standform equation that looks like this:
(x-h)^2=4p(y-k)
or
(y-k)^2=4p(x-h)
so, start chugging:
-2y^2+x-4y+1=0
-2y^2-4y = -x-1
-2(y^2+2) = -x-1
-2(y^2+2y+1) = -x-1 -2(1)
-2(y+1)^2 = -x-3
(y+1)^2 = 1/2(x+3)
Thank you so much!
The STD form of a parabola that opens to the right.
X = 2y^2 + 4y - 1.
Knock knock!
Who’s there?
Standard form.
Standard form who?
Standard form of the equation -2y^2 + x - 4y + 1 = 0!
To write the equation -2y^2 + x - 4y + 1 = 0 in standard form, we need to make sure that the equation is in the form Ax + By + C = 0, where A, B, and C are constants.
Step 1: Reorder the terms
Let's rearrange the equation so that the terms are in a more logical order:
x - 4y - 2y^2 + 1 = 0
Step 2: Group like terms
Next, let's group the terms with similar variables together:
x + (- 4y - 2y^2) + 1 = 0
Step 3: Rewrite the quadratic term
To rewrite the quadratic term in a standard form, we can factor out the greatest common factor from the quadratic coefficient:
x - (2y^2 + 4y) + 1 = 0
Step 4: Factor out a 2 from the terms with y
Now, let's factor out a 2 from the terms with y:
x - 2(y^2 + 2y) + 1 = 0
Step 5: Complete the square
To complete the square for the terms with y, we need to find a value that, when added and subtracted to the expression, does not change the equation:
x - 2(y^2 + 2y + 1) + 1 + 2(1) = 0
After adding and subtracting 1 inside the parentheses and 2 outside the parentheses, the equation is still equivalent.
Simplifying further, we get:
x - 2(y + 1)^2 + 3 = 0
Step 6: Rewrite the equation in standard form
Now that we have the equation in the desired form, we can rewrite it in standard form:
x - 2(y + 1)^2 + 3 = 0
Therefore, the equation -2y^2 + x - 4y + 1 = 0 can be rewritten in standard form as x - 2(y + 1)^2 + 3 = 0.