Question

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

y=−4x2+4x−8

(1 point)
Responses

y=−4(x−12)2−7
and (12,−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
and (12,7)
y equals negative 4 left parenthesis x minus Start Fraction 1 over 2 End Fraction right parenthesis squared plus 7 and left parenthesis Start Fraction 1 over 2 End Fraction comma 7 right parenthesis

y=−4(x−12)2−7
and (−12,−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
and (12,−7)

Answers

GPT-4o mini
To transform the quadratic equation from standard form to vertex form, we start with the given equation:

\[ y = -4x^2 + 4x - 8 \]

**Step 1: Factor out the coefficient of \( x^2 \)**

First, we factor out -4 from the first two terms:

\[ y = -4(x^2 - x) - 8 \]

**Step 2: Complete the square**

Next, we complete the square for the expression inside the parentheses. To do this, we take the coefficient of \( x \) (which is -1), halve it to get -0.5, and then square it to get 0.25. We add and subtract this inside the parentheses:

\[ y = -4\left(x^2 - x + 0.25 - 0.25\right) - 8 \]
\[ y = -4\left((x - 0.5)^2 - 0.25\right) - 8 \]

**Step 3: Expand and simplify**

Now distribute the -4:

\[ y = -4(x - 0.5)^2 + 1 - 8 \]
\[ y = -4(x - 0.5)^2 - 7 \]

So now we have the vertex form of the equation:

\[ y = -4(x - 0.5)^2 - 7 \]

**Step 4: Identify the vertex**

The vertex form \( y = a(x - h)^2 + k \) shows that the vertex of the parabola is at the point \( (h, k) \). Here, \( h = 0.5 \) and \( k = -7 \).

**Vertex point**: \( \left( \frac{1}{2}, -7 \right) \)

Thus, the final answer is:

\[ \boxed{y = -4\left(x - \frac{1}{2}\right)^2 - 7 \text{ and } \left( \frac{1}{2}, -7 \right)} \]