Problem 50 - Algebra for College Students

Question

Find the vertex and intercepts for each quadratic function. Sketch the graph, and state the domain and range. $$h(x)=-x^{2}-2 x+8$$

Step-by-Step Solution

Answer

Vertex: (-1, 9). X-intercepts: (-4, 0) and (2, 0). Y-intercept: (0, 8). Domain: (-∞, ∞). Range: (-∞, 9].

Step 1: Convert to Vertex Form

Rewrite the quadratic function in vertex form by completing the square. The given function is h(x)=-x^2-2x+8.Start with the quadratic and linear terms: -x^2 - 2x.Factor out the coefficient of the quadratic term:h(x) = - (x^2 + 2x) + 8.To complete the square inside the parentheses, add and subtract (2/2)^2 = 1:h(x) = - (x^2 + 2x + 1 - 1) + 8.Rewrite the expression inside the parentheses as a square:h(x) = - ((x + 1)^2 - 1) + 8.Simplify the expression:h(x) = - (x + 1)^2 + 1 + 8.So,h(x) = - (x + 1)^2 + 9.The vertex form of the function is - (x + 1)^2 + 9.

Step 2: Identify the Vertex

In vertex form, h(x) = - (x + 1)^2 + 9, the vertex can be identified as (h, k) = (-1, 9).So, the vertex is (-1, 9).

Step 3: Find the X-Intercepts

To find the x-intercepts, set h(x) to 0 and solve for x:0 = - x^2 - 2x + 8.Rewrite as a standard quadratic equation -x^2 - 2x + 8 = 0.Solve using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.Here, a = -1, b = -2, c = 8,so, x = [2 ± sqrt((-2)2 - 4(-1)(8))] / 2(-1) = [2 ± sqrt(4+32)] / -2 = [2 ± sqrt(36)] / -2 = [2 ± 6] / -2.Thus, x = (2 + 6) / -2 = -4 or x = (2 - 6) / -2 = 2.So, the x-intercepts are (-4, 0) and (2, 0).

Step 4: Find the Y-Intercept

To find the y-intercept, set x to 0 and solve for h(x): h(x) = - (0)^2 - 2(0) + 8 = 8.So, the y-intercept is (0, 8).

Step 5: Sketch the Graph

Plot the vertex (-1, 9),x-intercepts (-4, 0) and (2, 0),and y-intercept (0, 8).Since the quadratic opens downwards (the coefficient of x^2 is negative), draw a parabola opening downwards through these points.

Step 6: State the Domain and Range

The domain of any quadratic function is (-∞, ∞).For the range, since the parabola opens downwards and the vertex is the maximum point, the range is (-∞, 9].

Key Concepts

vertex form

When studying quadratic functions, rewriting the function in vertex form makes it easier to understand its geometric properties. The vertex form of a quadratic equation is given by:
\text{... Continue reading below if this helped:} \[ h(x) = a(x - h)^2 + k \]
Here, (h, k) represents the vertex of the parabola.
So how do we convert a quadratic function like \( h(x) = -x^2 - 2x + 8 \) into vertex form?
We complete the square!
First, start by factoring out the coefficient of the quadratic term:
\( h(x) = -(x^2 + 2x) + 8 \).
Next, add and subtract the square of half the linear term coefficient inside the parentheses: \( h(x) = -(x^2 + 2x + 1 - 1) + 8 \).
This allows us to rewrite the quadratic expression as a perfect square: \( h(x) = -((x + 1)^2 - 1) + 8 \).
Finally, simplify: \( h(x) = -(x + 1)^2 + 9 \).
Now we have the vertex form: \( h(x) = -(x + 1)^2 + 9 \).
From this, the vertex is clearly (-1, 9).

x-intercepts

To find the x-intercepts, we need to know where the function crosses the x-axis. This happens when \( h(x) = 0 \).
For the quadratic function \( h(x) = -x^2 - 2x + 8 \), set it to zero and solve for x: \( -x^2 - 2x + 8 = 0 \).
Applying the quadratic formula, which is \[ x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \], with \( a = -1 \), \( b = -2 \), and \( c = 8 \): \( x = \frac{2 \pm \sqrt{4 + 32}}{-2} \).
Simplify to get: \( x = \frac{2 \pm \sqrt{36}}{-2} \).
This translates to \( x = -4\) or \( x = 2\). So, the x-intercepts are (-4, 0) and (2, 0).

y-intercept

To find the y-intercept, identify where the function crosses the y-axis. This happens when \( x = 0 \).
For our quadratic function \( h(x) = -x^2 - 2x + 8 \): \( h(0) = -(0)^2 - 2(0) + 8 \).
So, \( h(0) = 8 \). Thus, the y-intercept is (0, 8). This is the point where the parabola intersects the y-axis.

completing the square

Completing the square is a method used to convert a quadratic function into vertex form. This makes it easier to identify the vertex and other key features of the graph.
To complete the square for \( h(x) = -x^2 - 2x + 8 \), follow these steps:

• Factor out the leading coefficient from the quadratic and linear terms: \( h(x) = -(x^2 + 2x) + 8 \).


• Add and subtract the square of half the coefficient of the linear term inside the parentheses: \( h(x) = -(x^2 + 2x + 1 - 1) + 8 \).


• Rewrite the quadratic expression inside the parentheses as a perfect square trinomial: \( h(x) = -((x + 1)^2 - 1) + 8 \).


• Simplify the equation: \( h(x) = -(x + 1)^2 + 9 \).

This gives us the vertex form of the function.

domain and range

The domain of any quadratic function is all real numbers: \( (-\infty, \infty) \). This is because we can input any real number value for x.
The range, however, depends on the vertex and the direction the parabola opens.
For the function \( h(x) = -(x + 1)^2 + 9 \), the parabola opens downward (as the coefficient of \( x^2 \) is negative), and the vertex is the highest point of the parabola.
Therefore, all output values (y-values) are less than or equal to the y-coordinate of the vertex, which is 9.
Thus, the range is: \( (-\infty, 9] \).