Problem 9 - Contemporary Precalculus

Question

In Exercises \(7-12,\) find the graph of the equation in the standard window. $$y=x^{2}-5 x+2$$

Step-by-Step Solution

Answer

Answer: The key features of the parabolic function $$y = x^2 - 5x + 2$$ are - Vertex: $$(\frac{5}{2},\frac{-9}{4})$$ - Axis of symmetry: $$x = \frac{5}{2}$$ - X-intercepts: $$x = \frac{5 + \sqrt{17}}{2}$$ and $$x = \frac{5 - \sqrt{17}}{2}$$ - Y-intercept: $$(0, 2)$$. To graph the function in the standard window, plot the vertex, x-intercepts, y-intercept, and axis of symmetry on a graph with ranges on both axes from -10 to 10. Sketch the parabola using these key features, and label the key points and axis of symmetry.

Step 1: 1. Determine the quadratic function form

The given equation is a quadratic function in the form $$y = ax^2 + bx + c$$, where $$a=1$$, $$b=-5$$, and $$c=2$$.

Step 2: 2. Find the vertex

The vertex of a parabolic function is given by the formula $$h = \frac{-b}{2a}$$. In this case, $$h = \frac{-(-5)}{2(1)} = \frac{5}{2}$$. Next, find the y-coordinate by plugging the $$x$$ value back into the equation: $$k = 1\left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 2$$ which simplifies to $$k = -\frac{9}{4}$$. Thus, the vertex of the parabola is $$(\frac{5}{2},\frac{-9}{4})$$.

Step 3: 3. Determine the axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. Its equation is given by $$x = h$$, where $$h$$ is the x-coordinate of the vertex. In this case, the axis of symmetry is $$x = \frac{5}{2}$$.

Step 4: 4. Find the x-intercepts

To find the x-intercepts, set $$y = 0$$ and solve the equation for $$x$$: $$0 = x^2 - 5x + 2$$ Using the quadratic formula, $$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$$, we have: $$x = \frac{5 ± \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 ± \sqrt{17}}{2}$$ So the x-intercepts are $$x = \frac{5 + \sqrt{17}}{2}$$ and $$x = \frac{5 - \sqrt{17}}{2}$$.

Step 5: 5. Find the y-intercept

To find the y-intercept, set $$x = 0$$ and solve the equation for $$y$$: $$y = 1(0)^2 - 5(0) + 2$$ $$y = 2$$ So the y-intercept is at the point $$(0, 2)$$.

Step 6: 6. Graph the parabola

Using the key features we've determined (vertex, x-intercepts, y-intercept, and the axis of symmetry) on a graph where the range on both axes is from -10 to 10, sketch the parabola and label the key points and axis of symmetry to find the graph of the equation in the standard window.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point that acts as either the maximum or minimum of the quadratic function, depending on the direction in which the parabola opens. For a parabola that opens upwards or downwards, the vertex is its highest or lowest point, respectively.
The vertex can be calculated using the formula \(h = \frac{-b}{2a}\), where \(a\) and \(b\) are coefficients from the quadratic equation \(y = ax^2 + bx + c\).

• In the exercise given, with \(a = 1\) and \(b = -5\), substituting these values gives \(h = \frac{-(-5)}{2(1)} = \frac{5}{2}\).
• To find the \(y\) coordinate of the vertex, substitute \(x = \frac{5}{2}\) back into the original equation: \(y = 1\left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 2\), simplifying to \(y = -\frac{9}{4}\).

Therefore, the vertex of the parabola \(y = x^2 - 5x + 2\) is located at \(\left(\frac{5}{2}, -\frac{9}{4}\right)\).

Axis of Symmetry

The axis of symmetry of a parabola is an imaginary line that runs through the vertex, dividing the parabola into two mirror images. It is an important feature that helps in analyzing the symmetry of the graph. For parabolas given by quadratic functions of the form \(y = ax^2 + bx + c\), the axis of symmetry can be derived from the solution for \(h\) in the vertex formula:

• For our specific equation, \(h = \frac{5}{2}\), so the axis of symmetry is the vertical line \(x = \frac{5}{2}\).

This means that if you were to fold the parabola at \(x = \frac{5}{2}\), both halves of the parabola would lay perfectly on top of each other.
This line not only passes through the vertex but also directs the path of the parabola's symmetry.

X-intercepts

The x-intercepts of a parabola are the points where the graph crosses the x-axis. These points are crucial as they represent the roots of the quadratic equation. To find these intercepts, set \(y = 0\) and solve the quadratic equation. This can be done using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• For the equation \(0 = x^2 - 5x + 2\), plug in \(a = 1\), \(b = -5\), and \(c = 2\) into the formula.
• This results in \(x = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}\).

Hence, the x-intercepts are \(x = \frac{5 + \sqrt{17}}{2}\) and \(x = \frac{5 - \sqrt{17}}{2}\).
These points help us understand at what x-values the parabola completely intersects the x-axis.

Y-intercept

A y-intercept is a point where the graph of a function crosses the y-axis. In the context of quadratic functions, this occurs by substituting \(x = 0\) into the function's equation. This provides a simple way to find one of the key points on the graph.

• For the equation \(y = x^2 - 5x + 2\), substituting \(x = 0\) gives \(y = 1(0)^2 - 5(0) + 2\).
• Simplifying this, we find \(y = 2\).

Thus, the y-intercept of the parabola is the point \((0, 2)\).
The y-intercept provides an initial point on the parabola, helping us to sketch the graph more accurately.