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Chapter 3: Problem 28

a. State whether the graph of the parabola b. Determine the vertex of the parabola. opens upward or downward. c. Determine the \(x\) -intercept(s). d. Determine the \(y\) -intercept. e. Sketch the graph. f. Determine the axis of symmetry. g. Determine the minimum or maximum h. State the domain and range. (See Example 3\()\) value of the function. $$ h(x)=-x^{2}-6 x-10 $$

Short Answer

Expert verified
The parabola opens downward. The vertex is (3, -37). The y-intercept is (0, -10). The axis of symmetry is x = 3. The maximum value is -37. Domain: all real numbers. Range: (-鈭�, -37].

Step by step solution

01

Determine the direction the parabola opens

Examine the coefficient of the quadratic term in the given equation. The equation is \(h(x) = -x^{2} - 6x - 10\). Since the coefficient of \(x^{2}\) is negative (-1), the parabola opens downward.
02

Find the vertex of the parabola

The vertex form of a parabola is \(h(x) = a(x-h)^2 + k\). For a standard form equation like \(ax^{2} + bx + c\), the vertex is at \(x = -\frac{b}{2a}\). Here, \(a = -1\) and \(b = -6\). So, \(x = -\frac{-6}{2(-1)} = 3\). Now, plug \(x = 3\) back into the equation to find \(h(x)\): \(h(3) = -(3)^2 - 6(3) - 10 = -9 - 18 - 10 = -37\). Thus, the vertex is at (3, -37).
03

Determine the x-intercepts

The x-intercepts are found by solving \(-x^{2} - 6x - 10 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here \(a = -1\), \(b = -6\), and \(c = -10\). So, \(x = \frac{6 \pm \sqrt{(-6)^2 - 4(-1)(-10)}}{2(-1)} = \frac{6 \pm \sqrt{36 - 40}}{-2} = \frac{6 \pm \sqrt{-4}}{-2} = \frac{6 \pm 2i}{-2}\). Therefore, the x-intercepts are complex numbers: \(x = 3 - i\) and \(x = 3 + i\).
04

Determine the y-intercept

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into \(h(x) = -0^{2} - 6(0) - 10 = -10\). So, the y-intercept is (0, -10).
05

Draw the axis of symmetry

The axis of symmetry of a parabola \(ax^{2} + bx + c\) is \(x = -\frac{b}{2a}\). For the given equation \(-x^{2} - 6x - 10\), the axis of symmetry is \(x = 3\).
06

Find the minimum or maximum value

Since the parabola opens downward, the vertex represents the maximum value. The maximum value of the function is \(-37\) at \(x = 3\).
07

State the domain and range

The domain of \(h(x) = -x^{2} - 6x - 10\) is all real numbers, because a parabola exists for all values of \(x\). The range of the function is \(-\infty < h(x) \leq -37\), since the maximum value is \(-37\).
08

Sketch the graph

Draw the axis of symmetry at \(x = 3\), plot the vertex at (3, -37), and the y-intercept at (0, -10). Since the roots are complex numbers, these are not shown on the real-number graph. Sketch the downward-opening shape based on these points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vertex of a parabola
The vertex of a parabola is a key feature that tells us its highest or lowest point. For a quadratic function in the form of \(ax^2 + bx + c\), the vertex can be found using the formula \(x = -\frac{b}{2a}\). Once we have the \(x\)-coordinate, we can substitute it back into the function to get the \(y\)-coordinate. For instance, in the equation \(h(x) = -x^2 - 6x - 10\), \(a = -1\) and \(b = -6\). Thus, \(x = -\frac{-6}{2(-1)} = 3\). Plugging \(x = 3\) back into the equation: \(h(3) = -(3)^2 - 6(3) - 10 = -37\). Hence, the vertex of this parabola is at \((3, -37)\).
x-intercepts
The x-intercepts of a parabola are points where the graph intersects the x-axis. These are found by solving the quadratic equation \(ax^2 + bx + c = 0\). In our example, \(h(x) = -x^2 - 6x - 10\), we use the quadratic formula: \(x = \frac{-b \textpm \riel-santhonlcaltinosaltb}{2a}\). Here, \(a = -1\), \(b = -6\), and \(c = -10\), so: \(x = \frac{6 \textpm \riel-santhonlcaltinosaltpositivenegative(36 - 40)}{-2} = \frac{6 \textpm \riel-2}{-2} = \frac{6 \textpm 2i}{-2}\). This results in complex numbers: \(3 - i\) and \(3 + i\), indicating there are no real x-intercepts.
y-intercepts
The y-intercept is the point where the parabola crosses the y-axis. To find this, set \(x\) to 0 and solve for \(h(x)\) in the equation. For \(h(x) = -x^2 - 6x - 10\), substitute \(x = 0\): \(h(0) = -0^2 - 6(0) - 10 = -10\). So, the y-intercept is at \((0, -10)\). This tells us that at \(x = 0\), the value of \(h(x)\) is \(-10\).
axis of symmetry
The axis of symmetry of a parabola is a vertical line that divides it into two symmetric halves. For a quadratic equation in standard form \(ax^2 + bx + c\), this line can be found using the equation \(x = -\frac{b}{2a}\). In our example, \(h(x) = -x^2 - 6x - 10\), \(a = -1\) and \(b = -6\). Thus, the axis of symmetry is \(x = 3\). This means that the x-value of 3 is the centerline of the parabola.
maximum value
A parabola can have either a maximum or minimum value, depending on whether it opens upward or downward. Our equation \(h(x) = -x^2 - 6x - 10\) has a negative coefficient for \(x^2\), indicating that the parabola opens downward and thus, has a maximum value. This maximum value occurs at the vertex. Plugging the vertex \(x\) value (3) back into the function, we find \(h(3) = -37\). Thus, the maximum value of the function is \(-37\).
domain and range
The domain of a parabola is the set of all possible \(x\)-values for which the function is defined. For any quadratic function, the domain is all real numbers. Hence, for \(h(x) = -x^2 - 6x - 10\), the domain is all real numbers. The range, however, is the set of all possible \(y\)-values the function can take. Because the parabola opens downward and has a maximum value at \(-37\), the range is \(-\riejaSanthombcaltinosaltimity_underline37]\).

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