Question

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

y=−5x2+30x−35

(1 point)
Responses

y=−5(x−3)2+10
and (−3,10)
y is equal to negative 5 times open paren x minus 3 close paren squared plus 10 and open paren negative 3 comma 10 close paren

y=5(x−3)2+10
and (3,10)
y is equal to 5 times open paren x minus 3 close paren squared plus 10 and open paren 3 comma 10 close paren

y=−5(x−3)2+10
and (3,10)
y is equal to negative 5 times open paren x minus 3 close paren squared plus 10 and open paren 3 comma 10 close paren

y=−5(x−3)2−10
and (3,−10)

Answer 1

To convert the quadratic equation \( y = -5x^2 + 30x - 35 \) to vertex form, we can complete the square.

1. Start with the equation:
\[
y = -5x^2 + 30x - 35
\]

2. Factor out \(-5\) from the first two terms:
\[
y = -5(x^2 - 6x) - 35
\]

3. To complete the square for \(x^2 - 6x\), take half of the coefficient of \(x\) (which is \(-6\)), square it, and add inside the parentheses. Half of \(-6\) is \(-3\), and \((-3)^2 = 9\):
- Add and subtract \(9\) inside the parentheses:
\[
y = -5(x^2 - 6x + 9 - 9) - 35
\]
This simplifies to:
\[
y = -5((x - 3)^2 - 9) - 35
\]

4. Distribute \(-5\):
\[
y = -5(x - 3)^2 + 45 - 35
\]

5. Simplify:
\[
y = -5(x - 3)^2 + 10
\]

Now we have the vertex form of the quadratic equation:
\[
y = -5(x - 3)^2 + 10
\]

The vertex of the parabola is given by the point \((h, k)\) from the vertex form \(y = a(x - h)^2 + k\). Here, \(h = 3\) and \(k = 10\), so the vertex point is:
\[
(3, 10)
\]

Thus, the answer is:
\[
y = -5(x - 3)^2 + 10 \quad \text{and} \quad (3, 10)
\]