Problem 25 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(g(x)=3(x+2)^{2}+5\)

Step-by-Step Solution

Answer

The vertex of the quadratic function \(g(x) = 3(x+2)^2 + 5\) is at point (-2, 5), the axis of symmetry is the line x = -2, and the y-intercept is at point (0, 17). There are no x-intercepts, as the function never intersects the x-axis. To graph the function, plot the vertex, axis of symmetry, and y-intercept, and draw the parabola opening upwards and narrower than the standard parabola.

Step 1: Identify the standard form of the quadratic function

Given function is: \(g(x) = 3(x+2)^2 + 5\) This is already given in vertex form, i.e., \(f(x) = a(x-h)^2 + k\), where 'a' represents the parabola's orientation and width, and the vertex is at point (h, k).

Step 2: Determine the vertex of the parabola

Comparing the given function with vertex form: \(g(x) = 3(x+2)^2 + 5\) We can see that: a = 3, h = -2, and k = 5 So, the vertex coordinates are: (h, k) = (-2, 5)

Step 3: Find the axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. The equation for the axis of symmetry is given by: x = h Using the values of h from the vertex coordinates: x = -2

Step 4: Calculate the x-intercepts and y-intercepts

The x-intercept is the point where the function crosses the x-axis (i.e., when the value of g(x) is 0): 0 = 3(x+2)^2 + 5 However, solving for x in this case does not give us any real solutions because the quadratic function never intersects the x-axis. Thus, there are no x-intercepts. The y-intercept is the point where the function crosses the y-axis (i.e., when the value of x is 0): g(0) = 3(0+2)^2 + 5 g(0) = 3(2)^2 + 5 g(0) = 3(4) + 5 g(0) = 12 + 5 g(0) = 17 So, the y-intercept is at the point (0, 17).

Step 5: Graph the function

To graph the function, follow these steps: 1. Plot the vertex of the parabola on the coordinate plane at the point (-2, 5) 2. Draw the axis of symmetry as a vertical line passing through the vertex (x = -2) 3. Plot the y-intercept at the point (0, 17) 4. Since 'a' is positive (a = 3), the parabola will open upwards and be narrower than the standard parabola. 5. Draw the parabola using the plotted points and the axis of symmetry Now the function is well graphed and the requested properties of the given quadratic function are determined.

Key Concepts

Vertex Form

The vertex form of a quadratic function can make it much easier to graph and understand the function. When expressed in vertex form, a quadratic equation looks like:
\[ f(x) = a(x-h)^2 + k \]
Here, \(a\), \(h\), and \(k\) are constants. This representation forms a parabola on the coordinate plane.

• \(a\) controls the opening direction and width of the parabola. If \(a\) is positive, the parabola opens upward; if negative, it opens downward.
• \(h\) and \(k\) locate the vertex of the parabola, a key point of symmetry located at \((h, k)\).

In the given exercise, the quadratic function is already expressed in vertex form:
\[ g(x) = 3(x+2)^2 + 5 \]
From this, we can easily identify that the vertex is at \((-2, 5)\).
This shows how the vertex form simplifies extracting detailed information about the quadratic equation.

Axis of Symmetry

The axis of symmetry is an integral part of understanding a quadratic function. It is a vertical line that splits the parabola into two mirror images.
The equation for the axis of symmetry is simple: it goes through the vertex and is given by \(x = h\).
In our specific quadratic function:
\[ g(x) = 3(x+2)^2 + 5 \]
the vertex is \((-2, 5)\).
So, the axis of symmetry is \(x = -2\).

• This axis is crucial for graphing the parabola because it guides the reflection and balance of the graph.
• Every point on the left of this line has a corresponding point on the right, maintaining a perfect symmetry.

X-intercept

The x-intercept(s) of a quadratic function is the point where the graph crosses the x-axis.
This occurs when the value of the function \(g(x)\) is zero. In other words, we set
\[ 3(x+2)^2 + 5 = 0 \]
and solve for \(x\).
In this exercise, attempting to solve the equation for real x-values reveals that there are no solutions.

• This indicates that the parabola does not intersect the x-axis.
• This typically happens when the entire parabola is situated either above or below the x-axis, depending on the value constants like \(a\) and the vertex location.

For this function, the parabola is entirely above the x-axis, supported by the fact that the vertex \((h, k) = (-2, 5)\) and \(a = 3\), which is positive.

Y-intercept

Finding the y-intercept of a quadratic function is more straightforward than the x-intercept. This is the point where the graph crosses the y-axis, specifically when \(x = 0\).
Plugging \(x = 0\) into the quadratic function:
\[ g(0) = 3(0+2)^2 + 5 \]
\[ g(0) = 12 + 5 \]
\[ g(0) = 17 \]
This calculation shows that the y-intercept of the function is at the point \((0, 17)\).

• The y-intercept provides an essential point for graphing the function.
• It also aids in understanding where the parabola sits in relation to the y-axis.

In this instance, the parabola sits notably above the x-axis, but intersects the y-axis at a relatively high point due to the positive \(a\) multiplier and the vertex's high placement at \((h, k) = (-2, 5)\).