Problem 32 Find the vertex of the graph of ... [FREE SOLUTION]

Chapter 8: Problem 32

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples 1 through 4 . $$ \frac{1}{5} x^{2}+2 x+\frac{9}{5} $$

Short Answer

Expert verified

The vertex is (-5, -\frac{16}{5}). The graph opens upward. Intercepts: y-intercept (0, \frac{9}{5}), x-intercepts (-1, 0) and (-9, 0).

Step by step solution

Identify the Quadratic Function Components

The given quadratic function is $ f(x) = \frac{1}{5}x^2 + 2x + \frac{9}{5} $. It can be expressed in the standard form $ ax^2 + bx + c $ where $ a = \frac{1}{5} $, $ b = 2 $, and $ c = \frac{9}{5} $.

Determine the Direction of the Parabola

The coefficient $ a = \frac{1}{5} $ is positive, indicating that the parabola opens upwards.

Find the Vertex

The formula for the x-coordinate of the vertex in a quadratic function $ ax^2 + bx + c $ is $ x = -\frac{b}{2a} $. Substituting the values, we have $ x = -\frac{2}{2 \times \frac{1}{5}} = -5 $. Substitute $ x = -5 $ back into the function to find the y-coordinate: \[ f(-5) = \frac{1}{5}(-5)^2 + 2(-5) + \frac{9}{5} = \frac{25}{5} - 10 + \frac{9}{5} = 5 - 10 + \frac{9}{5} = -5 + \frac{9}{5} = -\frac{16}{5} \].Therefore, the vertex is $(-5, -\frac{16}{5})$.

Find the Y-Intercept

To find the y-intercept, set $ x = 0 $ in the function: \[ f(0) = \frac{1}{5}(0)^2 + 2(0) + \frac{9}{5} = \frac{9}{5} \]. Therefore, the y-intercept is $(0, \frac{9}{5})$.

Find the X-Intercept(s)

Set the quadratic equation equal to zero to find the x-intercepts: \[ \frac{1}{5}x^2 + 2x + \frac{9}{5} = 0 \]. Multiply through by 5 to clear the fractions: \[ x^2 + 10x + 9 = 0 \]. Using the quadratic formula, $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 10 $, $ c = 9 $: \[ x = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times 9}}{2 \times 1} \].\[ x = \frac{-10 \pm \sqrt{100 - 36}}{2} = \frac{-10 \pm \sqrt{64}}{2} = \frac{-10 \pm 8}{2} \].So, $ x = -1 $ or $ x = -9 $. The x-intercepts are $(-1, 0)$ and $(-9, 0)$.

Sketch the Graph

Plot the vertex $(-5, -\frac{16}{5})$, the y-intercept $(0, \frac{9}{5})$, and the x-intercepts $(-1, 0)$ and $(-9, 0)$ on a graph. Since the parabola opens upward, draw the U-shaped curve passing through these points. Ensure symmetry around the vertex.

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Key Concepts

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

Vertex of a Parabola

In any quadratic function represented in the form $ ax^2 + bx + c $, the vertex is a critical point as it gives both the optimal and minimal position along the parabola's axis. To find the vertex, we use the formula for the x-coordinate: $ x = -\frac{b}{2a} $. For the quadratic $ f(x) = \frac{1}{5}x^2 + 2x + \frac{9}{5} $, substituting the values of $ a $ and $ b $ results in $ x = -5 $. By substituting this back into the function, we find $ f(x) = -\frac{16}{5} $, therefore the vertex is $(-5, -\frac{16}{5})$. This point is where the parabola changes direction and is pivotal when plotting or analyzing the curve.

Parabola Direction

The direction in which a parabola opens largely depends on the coefficient of the quadratic term $ a $. If $ a > 0 $, the parabola opens upwards, forming a shape like a U. If $ a < 0 $, it opens downwards like an upside-down U. This behavior affects the type of extremum the parabola has, whether a minimum at the vertex for an upward opening or a maximum for a downward opening.

X-Intercepts

X-intercepts are where the graph intersects the x-axis, meaning the y-value is zero at these points. To find the x-intercepts of $ f(x) = \frac{1}{5}x^2 + 2x + \frac{9}{5} $, set the equation to zero: $ \frac{1}{5}x^2 + 2x + \frac{9}{5} = 0 $. Solve by first clearing fractions, then applying the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $. For this equation, the x-intercepts are $(-1, 0)$ and $(-9, 0)$, providing points where the curve crosses the x-axis. These intercepts are essential for understanding the overall shape and behavior of the parabola.

Y-Intercepts

The y-intercept of a quadratic function is the point where the graph crosses the y-axis, and this occurs when $ x = 0 $. To find it in the function $ f(x) = \frac{1}{5}x^2 + 2x + \frac{9}{5} $, substitute $ x = 0 $ which simplifies the equation to $ f(0) = \frac{9}{5} $. Hence, the y-intercept is $(0, \frac{9}{5})$. The y-intercept provides a starting point for sketching the graph and helps to verify the accuracy of the plotted parabola.

91影视

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples 1 through 4 . $$ \frac{1}{5} x^{2}+2 x+\frac{9}{5} $$

Short Answer

Step by step solution

Identify the Quadratic Function Components

Determine the Direction of the Parabola

Find the Vertex

Find the Y-Intercept

Find the X-Intercept(s)

Sketch the Graph

Key Concepts

Vertex of a Parabola

Parabola Direction

X-Intercepts

Y-Intercepts

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