Question

Find the vertex of \(f(x)=-\frac{1}{2} x^{2}+4 x+5,\) and determine if the graph is concave up or concave down.

Short answer

The vertex is at (4, 13) and the graph is concave down.

Step-by-step solution

Identify the coefficients

The given quadratic function is in the standard form: \[ f(x) = ax^2 + bx + c \]Here, \( a = -\frac{1}{2} \), \( b = 4 \), and \( c = 5 \).

Determine the vertex using the vertex formula

The vertex form of a parabola \( f(x) = ax^2 + bx + c \) is found by using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \]. Plugging in the values, we get: \[ x = -\frac{4}{2(-\frac{1}{2})} \]. Calculate this to find the x-coordinate.

Calculate the x-coordinate of the vertex

Evaluating the expression from the previous step: \[ x = -\frac{4}{2(-\frac{1}{2})} = -\frac{4}{-1} = 4 \]. So, the x-coordinate of the vertex is 4.

Find the y-coordinate of the vertex

To find the y-coordinate, substitute the x-coordinate back into the original function: \[ f(4) = -\frac{1}{2}(4)^2 + 4(4) + 5 \]. Calculate this value step by step.

Calculate the y-coordinate

Evaluating the expression: \[ f(4) = -\frac{1}{2}(16) + 16 + 5 = -8 + 16 + 5 = 13 \]. So, the y-coordinate of the vertex is 13. Therefore, the vertex is at (4, 13).

Determine concavity

The concavity of the parabola is determined by the coefficient \( a \). If \( a > 0 \), the parabola is concave up. If \( a < 0 \), the parabola is concave down. Here, \( a = -\frac{1}{2} < 0 \), so the parabola is concave down.

Key concepts

parabolic vertex calculation

The vertex of a parabola is a vital point. It represents the maximum or minimum value of the quadratic function, depending on its concavity. For a given quadratic function in the form \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\).
The step-by-step process includes:

• Identify the coefficients (a, b, and c).
• Use the vertex formula for the x-coordinate.
• Substitute this x-value back into the function to find the y-coordinate.

In the example \(f(x) = -\frac{1}{2} x^2 + 4 x + 5\), the coefficients are \(a = -\frac{1}{2}\), \(b = 4\), and \(c = 5\). Using these values:
\[ x = -\frac{4}{2(-\frac{1}{2})} = -\frac{4}{-1} = 4 \]
The x-coordinate of the vertex is 4. To find the y-coordinate, plug x = 4 back into the function:
\[ f(4) = -\frac{1}{2}(4)^2 + 4(4) + 5 = -8 + 16 + 5 = 13 \]
So, the vertex of the function is (4, 13).

concavity of quadratic functions

Concavity describes the direction in which a parabola opens. It determines whether the vertex is a maximum or minimum point.
It is determined by the sign of the coefficient \(a\) in the quadratic function \(f(x) = ax^2 + bx + c\).

• If \(a > 0\), the parabola is concave up (opens upwards) and the vertex is a minimum point.
• If \(a < 0\), the parabola is concave down (opens downwards) and the vertex is a maximum point.

In our example, \(a = -\frac{1}{2}\), which is less than zero. Hence, the parabola is concave down. This means the vertex at (4, 13) is the highest point on the graph.
Understanding concavity helps in drawing the graph and analyzing the function's behavior over different intervals.

standard form of quadratic equations

The standard form of a quadratic function is \(f(x) = ax^2 + bx + c\). This form is crucial for identifying the coefficients which help in various calculations like finding the vertex and determining concavity.
Here’s a breakdown of what each term represents:

• \(ax^2\) - The quadratic term where \(a\) determines the width and the direction of the parabola.
• \(bx\) - The linear term which affects the slope of the parabola.
• \(c\) - The constant term that represents the y-intercept (the point where the parabola crosses the y-axis).

In the example function \(f(x) = -\frac{1}{2} x^2 + 4 x + 5\), \(a = -\frac{1}{2}\), \(b = 4\), and \(c = 5\). Recognizing these coefficients allows us to apply formulas like the vertex formula:
\[ x = -\frac{4}{2(-\frac{1}{2})} = 4 \]
By becoming familiar with the standard form and its components, solving quadratic functions becomes a structured and manageable process.