Problem 33 For quadratic function, (a) use ... [FREE SOLUTION]

Chapter 3: Problem 33

For quadratic function, (a) use the vertex formula to find the coordinates of the vertex and (b) graph the function. Do not use a calculator. $$P(x)=2 x^{2}-2 x+1$$

Short Answer

Expert verified

Vertex is $\left(\frac{1}{2}, \frac{1}{2}\right)$; graph as an upward-opening parabola.

Step by step solution

Identify A, B, and C

The given quadratic function is $P(x) = 2x^2 - 2x + 1$. Identify the coefficients $A$, $B$, and $C$ from the quadratic equation $Ax^2 + Bx + C$. Here, $A = 2$, $B = -2$, and $C = 1$.

Use the Vertex Formula

For a quadratic function $Ax^2 + Bx + C$, the x-coordinate of the vertex is given by the formula $x = -\frac{B}{2A}$. Substitute the values of $B$ and $A$ to find $x$:\[x = -\frac{-2}{2 \times 2} = \frac{2}{4} = \frac{1}{2}\]

Calculate the Y-coordinate of the Vertex

Substitute $x = \frac{1}{2}$ back into the function $P(x) = 2x^2 - 2x + 1$ to find the y-coordinate:\[P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 1\]Calculate each term:$2 \times \frac{1}{4} = \frac{1}{2}$, $-2 \times \frac{1}{2} = -1$Now, combine the terms: \[\frac{1}{2} - 1 + 1 = \frac{1}{2}\]Thus, the y-coordinate is $\frac{1}{2}$.

Determine the Vertex Coordinates

The vertex of the function $P(x)$ is at the point $\left(\frac{1}{2}, \frac{1}{2}\right)$.

Graph the Function

To graph $P(x) = 2x^2 - 2x + 1$, start by plotting the vertex at $\left(\frac{1}{2}, \frac{1}{2}\right)$. Since $A = 2$ is positive, the parabola opens upwards. Next, find additional points by selecting x-values (e.g., 0 or 1) to identify symmetry and shape. Calculate:\[P(0) = 2(0)^2 - 2(0) + 1 = 1\quad \text{(point (0, 1))}\]\[P(1) = 2(1)^2 - 2(1) + 1 = 1\quad \text{(point (1, 1))}\] These points confirm the parabola opens upwards with vertex $\left(\frac{1}{2}, \frac{1}{2}\right)$.

Draw the Parabola

Use the vertex and additional points ((0, 1) and (1, 1)) to sketch the parabola. It is symmetrical about the line $x = \frac{1}{2}$ and passes through the calculated points. Ensure it opens upwards.

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Key Concepts

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

Vertex Formula

The vertex formula is a key component when dealing with quadratic functions. It helps us locate the vertex, or the turning point, of a graph. For any quadratic function of the form $Ax^2 + Bx + C$, the x-coordinate of the vertex can be found using the formula $x = -\frac{B}{2A}$. This comes from completing the square or using calculus.

Identify the coefficients: In the equation $P(x) = 2x^2 - 2x + 1$, $A = 2$, $B = -2$, and $C = 1$.
Substitute into the formula to find $x$: $-\frac{-2}{2 \times 2} = \frac{1}{2}$.
Substitute $x = \frac{1}{2}$ back into the function to find $y$: $P(\frac{1}{2}) = \frac{1}{2}$.

Thus, the vertex of this quadratic function is located at $\ \left(\frac{1}{2}, \frac{1}{2}\right)$. This point is a critical component in understanding and graphing the function.

Graphing Quadratic Functions

Graphing quadratic functions involves several steps, starting with understanding the position and orientation of the graph. For the function $P(x) = 2x^2 - 2x + 1$, plotting the vertex is the starting point.

Place the vertex: Begin by graphing the vertex $\left(\frac{1}{2}, \frac{1}{2}\right)$.
Determine the direction: Since $A = 2$ is positive, the parabola opens upwards.
Find additional points: Select nearby x-values (e.g., $x = 0$ and $x = 1$) and calculate corresponding y-values.

For example, $P(0) = 1$ yields point (0,1), and $P(1) = 1$ yields point (1,1). These additional points help confirm the shape of the parabola. The graph will be symmetric about the line $x = \frac{1}{2}$, which runs through the vertex. Sketch the parabola ensuring its arms extend upwards.

Parabolas

Parabolas are the graph of quadratic functions and have several important properties. A parabola can open upwards or downwards, depending on the sign of the coefficient $A$ in the quadratic function.

If $A > 0$, the parabola opens upwards like a smile.
If $A < 0$, it opens downwards like a frown.
The vertex is the highest or lowest point on the parabola, depending on its direction.

A key feature of parabolas is their symmetry, which is about a vertical line passing through the vertex called the "axis of symmetry". For our function $P(x) = 2x^2 - 2x + 1$, the axis of symmetry is the line $x = \frac{1}{2}$. Parabolas appear frequently in different contexts, from physics (projectile motion) to economics (profit graphs). Understanding their properties helps navigate these real-world applications.

91影视

For quadratic function, (a) use the vertex formula to find the coordinates of the vertex and (b) graph the function. Do not use a calculator. $$P(x)=2 x^{2}-2 x+1$$

Short Answer

Step by step solution

Identify A, B, and C

Use the Vertex Formula

Calculate the Y-coordinate of the Vertex

Determine the Vertex Coordinates

Graph the Function

Draw the Parabola

Key Concepts

Vertex Formula

Graphing Quadratic Functions

Parabolas

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