Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Analyze the graph of the function algebraically and use the results to sketch the graph by hand. Then use a graphing utility to confirm your sketch. $$g(x)=-x^{2}+10 x-16$$

Step-by-Step Solution

Answer

The vertex of the parabola is at \((5, 9)\), y-intercept at \((0, -16)\), and roots at \((2,0)\) and \((8,0)\). Using these key points, the sketched graph should be an upside-down parabola, confirmed by the graphing utility.

Step 1: Determine the Vertex

The vertex form of a quadratic function is \(y=a(x-h)^{2}+k\), where \((h,k)\) is the vertex of the parabola. To convert the given function into vertex form, complete the square. The vertex of \(g(x)=-x^{2}+10x-16\) is \(\left(\frac{10}{2}, -\frac{(-100+64)}{4}\right)=\left(5, 9\right)\)

Step 2: Calculate the y-intercept

The y-intercept of the function is the value of \(y\) when \(x=0\), in this case, \(g(0)=-16\), so the y-intercept is at \((0,-16)\)

Step 3: Find the Roots

The roots of the function are calculated when \(g(x)=0\), which can be solved with the quadratic formula \(x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}\). For our function, a=-1, b=10, c=-16, thus the roots will be \(x=\frac{-10±\sqrt{10^{2}-4(-1)(-16)}}{2(-1)}=2\) and \(x=8\). Hence, the roots are at \((2,0)\) and \((8,0)\)

Step 4: Sketch the Graph

Using the vertex, y-intercept, and roots, draw an approximate sketch of the function on graph paper. Start by plotting the vertex \((5,9)\), the y-intercept \((0,-16)\), and the roots \((2,0)\) and \((8,0)\). Then connect these points with a smooth curve forming a parabola

Step 5: Confirm the Graph Using a Graphing Utility

Use a graphing utility (like Desmos, GeoGebra or a graphing calculator) to plot the function \(g(x)=-x^{2}+10x-16\). Make sure the digital graph aligns with the manual sketch.

Key Concepts

Vertex Form of a Quadratic

Understanding the vertex form of a quadratic function is crucial when you're analyzing graphs algebraically. This form is given by the equation \(y=a(x-h)^{2}+k\), where \(a\) is the coefficient that indicates the direction and width of the parabola, and \(h\) and \(k\) represent the coordinates of the vertex, the highest or lowest point on the graph.

Converting a quadratic equation to vertex form requires a process known as completing the square, which reveals the vertex. For the given function \(g(x)=-x^{2}+10 x-16\), completing the square helps us find the neat vertex \(\left(5, 9\right)\), the point where the parabola changes direction. This form is especially helpful in sketching the graph as it gives you a clear starting point.

Finding Roots Using Quadratic Formula

Quadratic equations often need to be solved for their roots, the values of \(x\) at which the function crosses the \(x\)-axis. The quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) comes to the rescue when factoring is not straightforward. It works for any quadratic function and guarantees finding the roots where \(a\), \(b\), and \(c\) are the coefficients of the equation \(ax^{2}+bx+c=0\).

In our function \(g(x)=-x^{2}+10x-16\), we identify \(a=-1\), \(b=10\), and \(c=-16\) to plug into the quadratic formula. As a result, the roots are found to be at \(x=2\) and \(x=8\). These are the points where the parabola intersects the \(x\)-axis, proving vital for graphing the function.

Graphing Parabolas

Graphing parabolas is a visual interpretation of quadratic functions. Every quadratic equation graphs into a parabola, which is a U-shaped curve. The shape and position of the parabola are determined by the equation's coefficients and its vertex. After finding the vertex, roots, and \(y\)-intercept, plotting the parabola becomes a tracing exercise. You start at the vertex and make a soft curve through the roots, all while ensuring that the graph mirrors itself on either side of the vertex because parabolas are symmetrical.

With our extracted values for \(g(x)=-x^{2}+10x-16\), we plot the key points and draw a parabolic curve that dips downwards, due to the negative coefficient of \(x^{2}\), displaying the expected symmetry.

Completing the Square

Completing the square is a method used to rewrite a quadratic function so that you can easily derive its vertex form and subsequently identify the vertex itself. This technique involves adding and subtracting a particular value so that part of the equation forms a perfect square trinomial. For instance, for the quadratic function \(g(x)=-x^{2}+10x-16\), you would add and subtract \(\left(\frac{10}{2}\right)^{2}\), resulting in an equation that can be factored into \(y=a(x-h)^{2}+k\) form.

The practical benefit is that once the function is in vertex form, the rest of the graphing process becomes easier and more intuitive. This strategy was utilized to find our vertex \(\left(5, 9\right)\) earlier in the solution.

Y-intercept of a Function

The \(y\)-intercept of a function is the point where the graph crosses the \(y\)-axis. It is found by setting \(x\) to zero and solving for \(y\). This intercept provides a clear starting or ending point when sketching the graph of an equation. For our function \(g(x)=-x^{2}+10x-16\), plugging in \(x = 0\) yields the \(y\)-intercept of \(\left(0, -16\right)\). It is an essential part of the graphing puzzle as it gives us another fixed point that the parabola will pass through.

In graphing exercises, spotting the \(y\)-intercept is a quick way to establish an anchor for your function on the coordinate plane, aiding in accuracy and giving confidence in plotting the curve.