Question

Transform the quadratic equation from standard form to vertex form and state the vertex point.

y=−4x2+4x−8
𝑦
=

4
𝑥
2
+
4
𝑥

8

(1 point)
Responses

y=−4(x−12)2−7
𝑦
=

4
(
𝑥

1
2
)
2

7
and (12,−7)
(
1
2
,

7
)
y equals negative 4 left parenthesis x minus Start Fraction 1 over 2 End Fraction right parenthesis squared minus 7 and left parenthesis Start Fraction 1 over 2 End Fraction comma negative 7 right parenthesis

y=−4(x−12)2−7
𝑦
=

4
(
𝑥

1
2
)
2

7
and (−12,−7)
(

1
2
,

7
)
y equals negative 4 left parenthesis x minus Start Fraction 1 over 2 End Fraction right parenthesis squared minus 7 and left parenthesis negative Start Fraction 1 over 2 End Fraction comma negative 7 right parenthesis

y=−4(x−12)2−7
𝑦
=

4
(
𝑥

1
2
)
2

7
and (12,−7)
(
1
2
,

7
)
y equals negative 4 left parenthesis x minus Start Fraction 1 over 2 End Fraction right parenthesis squared minus

Answers

GPT-4o mini
To transform the quadratic equation from standard form \( y = -4x^2 + 4x - 8 \) to vertex form \( y = a(x - h)^2 + k \) and to find the vertex point \( (h, k) \), we can use the method of completing the square.

1. Start with the standard form:
\[
y = -4x^2 + 4x - 8
\]

2. Factor out the coefficient of \( x^2 \) from the first two terms:
\[
y = -4\left(x^2 - x\right) - 8
\]

3. To complete the square, take the coefficient of \( x \) (which is -1), divide it by 2, and square it:
\[
\left(-\frac{1}{2}\right)^2 = \frac{1}{4}
\]

4. Add and subtract this square inside the parentheses:
\[
y = -4\left(x^2 - x + \frac{1}{4} - \frac{1}{4}\right) - 8
\]
\[
y = -4\left((x - \frac{1}{2})^2 - \frac{1}{4}\right) - 8
\]

5. Distribute the -4:
\[
y = -4\left(x - \frac{1}{2}\right)^2 + 1 - 8
\]
\[
y = -4\left(x - \frac{1}{2}\right)^2 - 7
\]

Now we can express the equation in vertex form and identify the vertex:

- The vertex form of the equation is:
\[
y = -4\left(x - \frac{1}{2}\right)^2 - 7
\]

- The vertex point \( (h, k) \) is:
\[
\left(\frac{1}{2}, -7\right)
\]

Therefore, the final answer is:
\[
y = -4\left(x - \frac{1}{2}\right)^2 - 7 \quad \text{and} \quad \left(\frac{1}{2}, -7\right)
\]