Problem 11 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(y=(x-4)^{2}-2\)

Step-by-Step Solution

Answer

The vertex of the quadratic function \(y=(x-4)^{2}-2\) is \((4, -2)\), the axis of symmetry is \(x=4\), the \(x\)-intercepts are \((4+\sqrt{2}, 0)\) and \((4-\sqrt{2}, 0)\), and the \(y\)-intercept is \((0, 14)\).

Step 1: Identify the vertex

The given function is in vertex form, which is \(y=a(x-h)^{2}+k\). The vertex is \((h, k)\), which can be found by comparing the given function with the vertex form. In the given function, we have \(h=4\) and \(k=-2\), so the vertex is \((4, -2)\).

Step 2: Identify the axis of symmetry

The axis of symmetry is a vertical line passing through the vertex, and its equation is of the form \(x=h\). In our case, since the vertex is \((4, -2)\), the axis of symmetry has equation \(x=4\).

Step 3: Find the \(x\)-intercepts

To find the \(x\)-intercepts, we set \(y=0\) and solve for x: \(0=(x-4)^{2}-2 \Rightarrow (x-4)^{2}=2\) Now, we need to solve for \(x\): \(x-4=\pm\sqrt{2}\) Adding 4 to both sides: \(x=4\pm\sqrt{2}\) There are two \(x\)-intercepts: \((4+\sqrt{2}, 0)\) and \((4-\sqrt{2}, 0)\).

Step 4: Find the \(y\)-intercept

To find the \(y\)-intercept, we set \(x=0\) and solve for \(y\): \(y=(0-4)^{2}-2=16-2=14\) The \(y\)-intercept is \((0, 14)\).

Step 5: Graph the function

To graph the function, first plot the vertex \((4, -2)\). Then, draw the axis of symmetry as a vertical line through the vertex, with equation \(x=4\). Next, plot the \(x\)-intercepts \((4+\sqrt{2}, 0)\) and \((4-\sqrt{2}, 0)\) and the \(y\)-intercept \((0, 14)\). Finally, sketch a parabola that is symmetric with respect to the axis of symmetry and passes through the vertex and the intercept points. The graph of the function \(y=(x-4)^{2}-2\) has a vertex at \((4, -2)\), axis of symmetry \(x=4\), \(x\)-intercepts \((4+\sqrt{2}, 0)\) and \((4-\sqrt{2}, 0)\), and a \(y\)-intercept \((0, 14)\).

Key Concepts

Vertex Form

The vertex form of a quadratic function is a powerful way to understand its properties immediately. The generic equation for vertex form is given by \[ y = a(x-h)^2 + k \]In this form,

• \( h \) and \( k \) represent the coordinates of the vertex of the parabola.
• \( a \) determines the direction and the width of the parabola.

For example, when we have a function like \( y = (x-4)^2 - 2 \), we can clearly see that the vertex

• is at the point \((h, k) = (4, -2)\).

This means the parabola opens upwards from the point \((4, -2)\). The vertex is a critical point since it represents the minimum or maximum value of the function depending on the direction of the parabola.

Axis of Symmetry

The axis of symmetry in a quadratic function gives us a vertical line that splits the parabola into two mirror-image halves. This line always passes through the vertex of the parabola. To find the equation for the axis of symmetry, you use the formula:

• \( x = h \), where \( h \) is the x-coordinate of the vertex.

In our example, \( y = (x-4)^2 - 2 \), the axis of symmetry

• has the equation \( x = 4 \).

This line tells us that the parabola mirrors itself over \( x = 4 \), providing a useful guide when sketching the curve.

X-Intercepts

The \( x \)-intercepts of a quadratic function are the points where the graph crosses the \( x \)-axis. To find these intercepts, set \( y = 0 \). For our quadratic \( y = (x-4)^2 - 2 \),

• we solve \( 0 = (x-4)^2 - 2 \).

Simplifying this:

• \((x-4)^2 = 2 \)
• \(x-4 = \pm \sqrt{2} \)

Thus, adding 4 to both solutions, we find

• \( x = 4 + \sqrt{2} \)
• \( x = 4 - \sqrt{2} \)

So, the \( x \)-intercepts are at

• \((4+\sqrt{2}, 0)\)
• \((4-\sqrt{2}, 0)\)

These intercepts help comprehensively map the parabola's path.

Y-Intercepts

Identifying the \( y \)-intercept is a fundamental step in graphing the quadratic function. It represents the point where the graph crosses the \( y \)-axis. To find the \( y \)-intercept, set \( x = 0 \) and solve for \( y \).For the function \( y = (x-4)^2 - 2 \), substituting \( x = 0 \) gives:

• \( y = (0-4)^2 - 2 \)
• \( y = 16 - 2 \)
• \( y = 14 \)

Thus, the \( y \)-intercept is at

• \((0, 14)\)

This point is crucial for constructing the full path of the parabola on a graph when combined with the vertex, axis of symmetry, and \( x \)-intercepts.