Problem 184 - Elementary Algebra

Question

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry. $$ y=-x^{2}+8 x-16 $$

Step-by-Step Solution

Answer

The x-intercept and vertex are (4, 0), y-intercept is (0, -16), and the axis of symmetry is x = 4.

Step 1 - Find the y-intercept

Set x = 0 in the equation y = -x^2 + 8x - 16 and solve for y . Since x = 0: y = - (0)^2 + 8(0) - 16 = -16. The y-intercept is (0, -16).

Step 2 - Find the x-intercepts

Set y = 0 in the equation: 0 = -x^2 + 8x - 16. Solve for x by factoring or using the quadratic formula. The equation can be rewritten as x^2 - 8x + 16 = 0. Factoring, we get (x-4)^2 = 0, so x = 4. The x-intercept is (4, 0).

Step 3 - Find the vertex

The vertex form of a parabola is given by y = a(x-h)^2 + k. Compare this with y = -x^2 + 8x - 16 to find h and k. Use the formula h = -b/2a to find the x-coordinate of the vertex. Here, a = -1 and b = 8, so h = -8/(2*(-1)) = 4. Substitute x = 4 back into the equation to find y: y = -4^2 + 8(4) - 16 = -16 + 32 - 16 = 0. Therefore, the vertex is (4, 0).

Step 4 - Find the axis of symmetry

The axis of symmetry of a parabola given in standard form y = ax^2 + bx + c is the vertical line that passes through the vertex. Since the vertex is (4, 0), the axis of symmetry is x = 4.

Step 5 - Graph the parabola

Use the points found — the y-intercept (0, -16), the x-intercept (4, 0), and the vertex (4, 0) — to plot the graph. Draw the axis of symmetry, x = 4, and sketch the parabola opening downwards.

Key Concepts

y-intercept

The y-intercept of a function is where the graph crosses the y-axis. To find it, we set x to 0 in the equation. In our example, given the quadratic equation \(y = -x^2 + 8x - 16\), we substitute x with 0:

\(y = - (0)^2 + 8(0) - 16\)
This simplifies to \(y = -16\). Therefore, the y-intercept is (0, -16). It tells us the graph touches the y-axis at this point.

x-intercepts

The x-intercepts are where the graph crosses the x-axis, meaning where y equals 0. For the given equation:
\(0 = -x^2 + 8x - 16\)
To find this, we set y to 0 and solve for x. Rewriting, we get:

\(x^2 - 8x + 16 = 0\).

We factor the quadratic to find:
\((x-4)^2 = 0\).
This means x = 4.

So, the x-intercept is at (4, 0). This point shows us where the graph crosses the x-axis.

vertex

The vertex of a quadratic function is the highest or lowest point of a parabola. It lies on the axis of symmetry. To find the vertex, we use the vertex form of a quadratic equation and compare it to our given equation:
\(y = -x^2 + 8x - 16\).
The formula for the x-coordinate of the vertex is \(h = -\frac{b}{2a}\). Here, \(a = -1\) and \(b = 8\). So,

\(h = \frac{-8}{2(-1)} = 4\).

We substitute x = 4 back into the equation to find y:
\(y = -4^2 + 8(4) - 16 = 0\).
Thus, the vertex is (4,0). This vertex also tells us that the maximum point of the graph is at (4,0).

axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola, dividing it into two mirror images. It's given by the equation \( x = h \). For our quadratic equation:
Since the vertex is at (4, 0), the axis of symmetry is:
\( x = 4 \).
The axis helps us understand how the parabola is symmetric and aids in graphing.

parabola

A parabola is the graph of a quadratic function and has a symmetrical U-shaped curve. It can either open upwards or downwards. For our equation \(y = -x^2 + 8x - 16\), the parabola opens downwards because the coefficient of \(x^2\) is negative.

To graph the parabola, we use the points we’ve found: the y-intercept (0, -16), the x-intercept (4, 0), and the vertex (4, 0). We also draw the axis of symmetry, x = 4. Using these points, we can sketch the parabolic curve.