Problem 7 - Precalculus Mathematics for Calculus

Question

The graph of a quadratic function \(f\) is given. (a) Find the coordinates of the vertex. (b) Find the maximum or minimum value of \(f\). (c) Find the domain and range of \(f\). $$f(x)=2 x^{2}-4 x-1$$ (GRAPH CAN'T COPY)

Step-by-Step Solution

Answer

Vertex: (1, -3), Minimum: -3, Domain: (-∞, ∞), Range: [-3, ∞).

Step 1: Finding the Vertex

The vertex form of a quadratic function is given by \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex. We can find \(h\) using the formula \(h = -\frac{b}{2a}\). For the given equation \(f(x) = 2x^2 - 4x - 1\), \(a = 2\) and \(b = -4\). Using the formula, \(h = -\frac{-4}{2 \times 2} = 1\). To find \(k\), substitute \(x = 1\) back into the function: \(f(1) = 2(1)^2 - 4(1) - 1 = -3\). Therefore, the vertex is \((1, -3)\).

Step 2: Identifying Maximum or Minimum Value

The maximum or minimum value of a quadratic function occurs at the vertex. Since the leading coefficient \(a = 2 > 0\), which means the parabola opens upwards, the function has a minimum value. Thus, the minimum value of the function is the \(y\)-coordinate of the vertex, which is \(-3\).

Step 3: Determining the Domain

The domain of a quadratic function is all real numbers because the function is defined for every real number \(x\). Therefore, the domain is \((-\infty, \infty)\).

Step 4: Determining the Range

Since the quadratic function opens upwards and the minimum value (vertex \(y\)-coordinate) is \(-3\), the range starts at \(-3\) and extends to infinity. Therefore, the range is \([-3, \infty)\).

Key Concepts

Vertex of a Parabola

In a quadratic function, the vertex is a key feature. It's the point where the parabola changes direction. Understanding the vertex helps us understand the parabola's behavior. To find the vertex from the quadratic function given in standard form, we first use the formula for the vertex

• Recall the quadratic function: \( ax^2 + bx + c \).


• The vertex formula helps find the \(x\)-coordinate: \( h = -\frac{b}{2a} \).

Then, plug this value back into the function to find the \(y\)-coordinate, \(k\).
For the quadratic function \( f(x) = 2x^2 - 4x - 1 \):

• Calculate \(h\): \( h = -\frac{-4}{2 \times 2} = 1 \).


• Substitute \(h = 1\) into \(f(x)\) to find \(k\): \(f(1) = 2(1)^2 - 4(1) - 1 = -3\).

The vertex is \((1, -3)\). This point is crucial as it reveals the peak or dip of the parabola.

Maximum and Minimum Values

Quadratic functions are interesting because they can have a maximum or a minimum value. This is found at the vertex of the parabola.
If the parabola opens upwards, it has a minimum value. If it opens downwards, it has a maximum.

• For our function \( f(x) = 2x^2 - 4x - 1 \), the coefficient \(a = 2\) is positive.


• This means the parabola opens upwards, indicating a minimum value.


• The minimum value is the \(y\)-coordinate of the vertex, which is \(-3\).

Understanding whether a parabola has a maximum or minimum helps in knowing the range of values the function can take. This property is particularly useful in optimization problems.

Domain and Range

The domain and range of quadratic functions describe all possible inputs and outputs of the function.
The domain of a quadratic function is always all real numbers.

• The function \(f(x)\) is defined for every real number \(x\).


• This means the domain is \(( -\infty , \infty )\).

For the range:

• It's determined by the direction the parabola opens.


• Since the parabola opens upwards and reaches a minimum at the vertex, the range starts at the vertex's \(y\)-coordinate \(-3\) and goes to infinity.

Thus, the range of \(f(x) = 2x^2 - 4x - 1\) is \([-3, \infty)\). Being able to calculate the domain and range allows us to fully understand the limits and possibilities of any quadratic function.