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Magazine Chapter 12: Problem 53
Graph each function using the vertex formula. Include the intercepts. \(h(x)=-\frac{1}{3} x^{2}-2 x-5\)
Short Answer
Expert verified
The vertex of \(h(x)=-\frac{1}{3}x^2-2x-5\) is at \((-3, -2)\), the x-intercepts are approximately at \((1.15, 0)\) and \((-4.48, 0)\), and the y-intercept is at \((0, -5)\).
Step by step solution
01
1. Find the vertex using the vertex formula
To find the vertex of the quadratic function \(h(x)=-\frac{1}{3}x^2-2x-5\), we can use the vertex formula:
\[x_V = \frac{-b}{2a}\]
where a and b are the coefficients of the \(x^2\) and \(x\) terms, respectively. In this case, \(a=-\frac{1}{3}\) and \(b=-2\). Plug these values into the formula to find the x-coordinate of the vertex:
\[x_V = \frac{-(-2)}{2(-\frac{1}{3})} = \frac{2}{-\frac{2}{3}} = -3\]
Now that we have the x-coordinate, we can find the y-coordinate of the vertex by plugging \(x_V\) into the function:
\[y_V = h(-3) = -\frac{1}{3}(-3)^2 - 2(-3) - 5 = -\frac{1}{3}(9) + 6 - 5 = -3 + 6 - 5 = -2\]
Therefore, the vertex is at the point \((-3, -2)\).
02
2. Find the x-intercepts (if any) by setting \(y=0\) and solving for '\(x\)'.
To find the x-intercepts, we need to find the values of '\(x\)' for which the function \(h(x)\) is equal to 0. Set \(h(x) = 0\) and solve for \(x\):
\[-\frac{1}{3}x^2 - 2x - 5 = 0\]
Unfortunately, this equation cannot be factored easily. Therefore, we will use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this case, \(a = -\frac{1}{3}\), \(b = -2\), and \(c = -5\). Plugging these values into the formula, we get:
\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-\frac{1}{3})(-5)}}{2(-\frac{1}{3})}\]
\[x = \frac{2 \pm \sqrt{4 + \frac{20}{3}}}{-\frac{2}{3}}\]
Evaluate the expression inside the square root:
\[\sqrt{4 + \frac{20}{3}} = \sqrt{\frac{12 + 20}{3}} = \sqrt{\frac{32}{3}}\]
Now plug this value back into the equation for '\(x\)':
\[x = \frac{2 \pm \sqrt{\frac{32}{3}}}{-\frac{2}{3}}\]
Because the values of '\(x\)' are not easy to calculate and involve irrational numbers, we will approximate the x-intercepts as \((1.15, 0)\) and \((-4.48, 0)\).
03
3. Determine the y-intercept by setting \(x = 0\) and solving for '\(y\)'.
To find the y-intercept, substitute \(x=0\) into the function and solve for \(h(x)\):
\[h(0) = -\frac{1}{3}(0)^2 - 2(0) - 5 = -5\]
Therefore, the y-intercept is at the point \((0, -5)\).
04
4. Plot and graph the function
Now, we can plot the vertex \((-3, -2)\), the x-intercepts \((1.15, 0)\) and \((-4.48, 0)\), and the y-intercept \((0, -5)\). Using these points as a guide, sketch the graph of the function, which should resemble a parabola opening downward.
Conclusion: We have graphed the given function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\) using the vertex formula and intercepts. The vertex is at \((-3, -2)\), the x-intercepts are approximately at \((1.15, 0)\) and \((-4.48, 0)\), and the y-intercept is at \((0, -5)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Formula
To find the vertex of a quadratic function, we use the vertex formula. This is an essential tool, especially for functions in standard form, like \( h(x) = ax^2 + bx + c \). The x-coordinate of the vertex \((x_V)\) is determined using the formula:
In our example, the function \(h(x)=-\frac{1}{3}x^2-2x-5\) yields the vertex at \((-3, -2)\), where \( x_V = -3 \) and \( y_V = -2 \). This vertex represents the highest point on the graph since the parabola opens downward.
- \( x_V = \frac{-b}{2a} \)
In our example, the function \(h(x)=-\frac{1}{3}x^2-2x-5\) yields the vertex at \((-3, -2)\), where \( x_V = -3 \) and \( y_V = -2 \). This vertex represents the highest point on the graph since the parabola opens downward.
X-Intercepts
Finding the x-intercepts involves determining where the graph crosses the x-axis. This is done by setting the function equal to zero and solving for \( x \). For a quadratic function \( ax^2 + bx + c = 0 \), the x-intercepts can be found using:
In the given function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\), solving with the quadratic formula gives approximate x-intercepts at \((1.15, 0)\) and \((-4.48, 0)\). These points are crucial for sketching an accurate graph, indicating where the graph touches or crosses the x-axis.
- Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In the given function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\), solving with the quadratic formula gives approximate x-intercepts at \((1.15, 0)\) and \((-4.48, 0)\). These points are crucial for sketching an accurate graph, indicating where the graph touches or crosses the x-axis.
Y-Intercepts
The y-intercept is where the graph intersects the y-axis. To find it, set \( x = 0 \) in the function, then solve for \( y \). This point is straightforward because it represents the value of the function when all \( x \)-terms are zero.
For the function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\), substitute \( x = 0 \) to find:
For the function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\), substitute \( x = 0 \) to find:
- \( h(0) = -5 \)
Graphing Parabolas
Graphing a parabola requires identifying critical points such as the vertex and intercepts. Once these are found, they provide a framework for sketching the curve. A quadratic function in standard form will graph as a parabola either opening upwards or downwards.
In our example, the function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\) presents a downward-opening parabola.
In our example, the function \(h(x) = -\frac{1}{3}x^2 - 2x - 5\) presents a downward-opening parabola.
- The vertex at \((-3, -2)\) shows the peak.
- X-intercepts at approximately \((1.15, 0)\) and \((-4.48, 0)\) indicate where the graph crosses the x-axis.
- The y-intercept at \((0, -5)\) shows where the graph crosses the y-axis.
