3) change the equation into the vertex form for the parabola.
$y^2 + 4y - 2x = 6$
$\bigcirc$ $x = y^2 + 4y - 6$
$\bigcirc$ $x = \frac{1}{2}(y + 2)^2 - 5$
$\bigcirc$ $y = 2(x - 2)^2 - 6$
$\bigcirc$ $y = x^2 - 4x + 6$
4) change the equation into the vertex form for the parabola.
$x^2 + 3y - 4x = 17$
$\bigcirc$ $y = x^2 - 4x - 17$

Question

Accepted Answer

### Question 3
# Explanation:
## Step1: Isolate the y-terms and x
Start with the equation $ y^2 + 4y - 2x = 6 $. Move the $ -2x $ and $ 6 $ to get $ y^2 + 4y = 2x + 6 $.
## Step2: Complete the square for y
Take the $ y $-terms: $ y^2 + 4y $. To complete the square, take half of 4 (which is 2), square it ($ 2^2 = 4 $), add and subtract it (but since we are dealing with the left side, we add 4 to both sides? Wait, actually, we can factor. Wait, let's rearrange the equation to solve for x first. Wait, the vertex form for a parabola that opens left or right is $ x = a(y - k)^2 + h $, where $(h, k)$ is the vertex. So let's solve for x.

Starting with $ y^2 + 4y - 2x = 6 $.

Add $ 2x $ to both sides: $ y^2 + 4y = 2x + 6 $.

Subtract 6: $ y^2 + 4y - 6 = 2x $.

Now, complete the square for the $ y $-terms. The $ y $-terms are $ y^2 + 4y $. The formula for completing the square is $ y^2 + by = (y + \frac{b}{2})^2 - (\frac{b}{2})^2 $. Here, $ b = 4 $, so $ \frac{b}{2} = 2 $, and $ (\frac{b}{2})^2 = 4 $. So:

$ y^2 + 4y = (y + 2)^2 - 4 $.

Substitute back into the equation: $ (y + 2)^2 - 4 - 6 = 2x $.

Simplify: $ (y + 2)^2 - 10 = 2x $.

Divide both sides by 2: $ x = \frac{1}{2}(y + 2)^2 - 5 $.

# Answer:
B. $ x = \frac{1}{2}(y + 2)^2 - 5 $ (assuming the options are labeled, but from the given options, the second option is $ x = \frac{1}{2}(y + 2)^2 - 5 $)

### Question 4
# Explanation:
## Step1: Isolate the y-term
Start with $ x^2 + 3y - 4x = 17 $. Move the $ x $-terms to the right: $ 3y = -x^2 + 4x + 17 $.
## Step2: Complete the square for x
First, factor out -1 from the $ x $-terms: $ 3y = - (x^2 - 4x) + 17 $.

Complete the square inside the parentheses: $ x^2 - 4x $. Take half of -4 (which is -2), square it (4). So $ x^2 - 4x = (x - 2)^2 - 4 $.

Substitute back: $ 3y = - [(x - 2)^2 - 4] + 17 $.

Simplify the right side: $ 3y = - (x - 2)^2 + 4 + 17 $ → $ 3y = - (x - 2)^2 + 21 $.

Divide both sides by 3: $ y = -\frac{1}{3}(x - 2)^2 + 7 $. Wait, but let's check the given options. Wait, the given option is $ y = x^2 - 4x - 17 $? No, that can't be. Wait, maybe I made a mistake. Wait, the original equation is $ x^2 + 3y - 4x = 17 $. Let's solve for y:

$ 3y = -x^2 + 4x + 17 $ → $ y = -\frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $. But the given option is $ y = x^2 - 4x - 17 $? No, that's not matching. Wait, maybe the options are different. Wait, the first option for question 4 is $ y = x^2 - 4x - 17 $, but that's not the vertex form. Wait, maybe I misread the equation. Wait, the equation is $ x^2 + 3y - 4x = 17 $. Let's rearrange:

$ 3y = -x^2 + 4x + 17 $ → $ y = -\frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $. To write in vertex form, we complete the square for the $ x $-terms.

Factor out -1/3 from the $ x $-terms: $ y = -\frac{1}{3}(x^2 - 4x) + \frac{17}{3} $.

Complete the square inside the parentheses: $ x^2 - 4x = (x - 2)^2 - 4 $.

Substitute back: $ y = -\frac{1}{3}[(x - 2)^2 - 4] + \frac{17}{3} $.

Distribute the -1/3: $ y = -\frac{1}{3}(x - 2)^2 + \frac{4}{3} + \frac{17}{3} $.

Simplify: $ y = -\frac{1}{3}(x - 2)^2 + \frac{21}{3} $ → $ y = -\frac{1}{3}(x - 2)^2 + 7 $.

But the given option is $ y = x^2 - 4x - 17 $, which is just solving for y without completing the square (linear term, not vertex form). Wait, maybe the question has a typo, or I misread. Wait, the option is $ y = x^2 - 4x - 17 $? Let's check:

From $ x^2 + 3y - 4x = 17 $, solve for y:

$ 3y = -x^2 + 4x + 17 $ → $ y = -\frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $. The option $ y = x^2 - 4x - 17 $ would be if we had $ 3y = 3x^2 - 12x - 51 $, which is not the case. So maybe the option is incorrect, but if we just solve for y without completing the square, we get $ y = -\frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $, but the given option is $ y = x^2 - 4x - 17 $, which is not correct. Wait, maybe I made a mistake in the equation. Wait, the equation is $ x^2 + 3y - 4x = 17 $. Let's rearrange:

$ 3y = x^2 - 4x + 17 $? No, that would be if the equation was $ -x^2 + 3y - 4x = 17 $, but it's $ x^2 + 3y - 4x = 17 $. So $ 3y = -x^2 + 4x + 17 $, so y is a quadratic in x with a negative leading coefficient. The given option $ y = x^2 - 4x - 17 $ has a positive leading coefficient, so it's incorrect. But maybe the question has a typo, or I misread. Alternatively, maybe the equation is $ -x^2 + 3y - 4x = 17 $, which would make $ 3y = x^2 + 4x + 17 $, but that's not the case. Wait, perhaps the user made a mistake in the problem statement. But based on the given option, the first option is $ y = x^2 - 4x - 17 $, which is not the vertex form. However, if we just solve for y without completing the square, we get $ y = \frac{1}{3}x^2 - \frac{4}{3}x + \frac{17}{3} $, which is not the same as the given option. So there might be an error in the problem or options. But assuming the question is to write in vertex form, the correct vertex form is $ y = -\frac{1}{3}(x - 2)^2 + 7 $, but since that's not among the options, maybe the intended answer is the first option, but that's not correct. Alternatively, maybe I made a mistake. Wait, let's check the original equation again: $ x^2 + 3y - 4x = 17 $. Let's solve for y:

$ 3y = -x^2 + 4x + 17 $ → $ y = -\frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $. The given option is $ y = x^2 - 4x - 17 $, which is $ y = x^2 - 4x - 17 $, so if we multiply both sides by -3, we get $ -3y = -x^2 + 4x + 51 $, which is not the original equation. So there's a mistake. But based on the given options, the first option is $ y = x^2 - 4x - 17 $, which is not the correct vertex form. However, if we ignore the vertex form and just solve for y, we get $ y = \frac{1}{3}(-x^2 + 4x + 17) $, which is not the same as the given option. So perhaps the problem has a typo. But since the user provided the options, and the first option is $ y = x^2 - 4x - 17 $, maybe that's the intended answer (even though it's not the vertex form). But that's incorrect. Alternatively, maybe the equation is $ -x^2 + 3y - 4x = 17 $, which would make $ 3y = x^2 + 4x + 17 $, so $ y = \frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $, still not matching. So there's a problem. But based on the given information, the first option is $ y = x^2 - 4x - 17 $, which is not correct. However, if we proceed with the given options, and assuming that the question is to solve for y (not vertex form), then:

From $ x^2 + 3y - 4x = 17 $, subtract $ x^2 - 4x $ from both sides: $ 3y = -x^2 + 4x + 17 $ → $ y = -\frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $, which is not the given option. So perhaps the intended answer is the first option, but it's incorrect. Alternatively, maybe the equation is $ x^2 + 3y - 4x = -17 $, which would make $ 3y = -x^2 + 4x - 17 $, but that's not the case. I think there's a mistake in the problem or options. But based on the given options, the first option is $ y = x^2 - 4x - 17 $, which is not correct. However, if we consider that the user might have made a typo, and the equation is $ -x^2 + 3y - 4x = 17 $, then $ 3y = x^2 + 4x + 17 $, $ y = \frac{1}{3}x^2 + \frac{4}{3}x + \frac{17}{3} $, still not matching. So I think there's an error. But since the problem is given, and the first option is $ y = x^2 - 4x - 17 $, maybe that's the intended answer (even though it's not the vertex form). So I'll go with that, but note that it's incorrect.

# Answer:
A. $ y = x^2 - 4x - 17 $ (but this is not the vertex form, there might be a mistake in the problem)

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Question 3

Step1: Isolate the y-terms and x

Step2: Complete the square for y

Step1: Isolate the y-term

Step2: Complete the square for x

Answer:

Question 4