Write the equation (in vertex form) of the parabola with the given characteristics:

Question

YOU MUST TYPE f, of, x, equalsf(x)= NOT y, equalsy=

leading coefficient: minus, 2−2

vertex left parenthesis, 5, comma, minus, 3, right parenthesis(5,−3)
Answer
Attempt 1 out of 3

equation:

vertex:

range:

Answer 1

To write the equation of a parabola in vertex form, we use the formula:

\[ f(x) = a(x - h)^2 + k \]

where \((h, k)\) is the vertex of the parabola, and \(a\) is the leading coefficient.

Given the characteristics:

• Leading coefficient: \(-2\)
• Vertex: \((5, -3)\)

We can substitute these values into the vertex form equation.

1. The vertex \((h, k)\) is \((5, -3)\), so \(h = 5\) and \(k = -3\).
2. The leading coefficient \(a\) is \(-2\).

Now we can substitute these values into the formula:

\[ f(x) = -2(x - 5)^2 - 3 \]

Answer:

Equation:
\[ f(x) = -2(x - 5)^2 - 3 \]

Vertex:
\((5, -3)\)

Range:

Since the parabola opens downwards (because \(a < 0\)), the vertex will be the maximum point. The maximum value of the function occurs at the vertex's y-value, which is \(-3\).

Therefore, the range of \(f(x)\) is: \[ (-\infty, -3] \]