Problem 28 - Algebra for College Students

Question

Graph each quadratic function, and state its domain and range. $$h(x)=(x+3)^{2}$$

Step-by-Step Solution

Answer

Domain: \( \mathbb{R} \); Range: \( [0, \infty) \)

Step 1: Rewrite the Function in Vertex Form

The given quadratic function is already in vertex form: \[ h(x) = (x + 3)^2 \]Here, the vertex form of a quadratic function is typically written as \[ y = a(x - h)^2 + k \]. In this case, it's \[ y = (x - (-3))^2 \].

Step 2: Identify the Vertex

The vertex of the quadratic function \( h(x) = (x + 3)^2 \) can be determined by identifying the values of \(-h\) and \(k\). Comparing with the vertex form, the vertex is at \((-3, 0)\).

Step 3: Determine the Direction of the Parabola

The coefficient in front of \((x + 3)^2\) is positive \( (a = 1) \). This means the parabola opens upwards.

Step 4: Find Intercepts

To find the y-intercept, set \( x = 0 \) and solve for \( y \): \[ h(0) = (0 + 3)^2 = 9 \]Thus, the y-intercept is \( (0, 9) \).For the x-intercepts, set \( y = 0 \) and solve for \( x \): \[ 0 = (x + 3)^2 \]Solving this, we find that \( x = -3 \).Thus, the x-intercept is \( (-3, 0) \).

Step 5: Draw the Graph

Plot the vertex \((-3, 0)\) and the intercepts. Sketch the parabola opening upwards with the vertex at \((-3, 0)\) and passing through the points reflecting the intercepts. Ensure that it opens symmetrically.

Step 6: State the Domain

The domain of any quadratic function is all real numbers. Thus, the domain of \( h(x) = (x + 3)^2 \) is \( \mathbb{R} \).

Step 7: State the Range

Since the parabola opens upwards and the vertex is the minimum point at \( y = 0 \), the range is all real numbers greater than or equal to 0. Thus, the range is: \[ [0, \infty) \].

Key Concepts

vertex form

The vertex form of a quadratic function is very handy for graphing and analyzing the function. The general vertex form is written as: \[ y = a(x - h)^2 + k \] Here, \(h\) and \(k\) represent the coordinates of the vertex of the parabola. In the function \[ h(x) = (x + 3)^2 \] we see that \(-3\) and \(0\) are the values, so the vertex is \((-3, 0)\). This is crucial because the vertex is the point where the parabola changes direction. The coefficient \(a\) in the vertex form determines how 'wide' or 'narrow' the parabola is and whether it opens upwards or downwards. Make sure to identify \(a\), \(h\), and \(k\) correctly to properly graph the function.

domain and range

Let's talk about the domain and range of quadratic functions.

• The domain of a quadratic function is always all real numbers because you can plug any number into a quadratic function and get a valid result. So, for \[ h(x) = (x + 3)^2 \], the domain is \[ \mathbb{R} \].
• The range depends on the direction the parabola opens and the vertex of the parabola. Since our function opens upward and has its vertex at \((-3, 0)\), the smallest value of \(y\) is 0. Thus, the range is \[ [0, \infty) \], which means all real numbers greater than or equal to 0.

parabola direction

The direction in which a parabola opens is determined by the coefficient \(a\) in the vertex form \[ y = a(x - h)^2 + k \].

• If \(a > 0\), the parabola opens upward.
• If \(a < 0\), the parabola opens downward.

In the function \[ h(x) = (x + 3)^2 \], \(a\) is 1, which is positive, indicating that the parabola opens upwards. This is crucial because it tells us that the vertex is the minimum point on the graph. This affects the range of the function as well.

intercepts

Identifying the intercepts of a quadratic function is very important for graphing.

• Y-intercept: This is where the graph crosses the y-axis. You find it by setting \(x = 0\) and solving for \(y\). For \[ h(x) = (x + 3)^2 \], substitute \(0\) for \(x\) to get \( h(0) = 9 \). So, the y-intercept is \((0, 9)\).
• X-intercepts: These are the points where the graph crosses the x-axis. You find it by setting \(y = 0\) and solving for \(x\). For \[ h(x) = (x + 3)^2 \], set \( h(x) = 0 \), so \[ 0 = (x + 3)^2 \], solving this gives \( x = -3 \). The x-intercept is \((-3, 0)\).

Plotting these points helps in sketching the accurate shape of the parabola, making it easier to understand and analyze the function.