Problem 17 - Precalculus

Question

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$y=x^{2}-2 x+1$$

Step-by-Step Solution

Answer

The graph is a parabola with vertex at (1, 0) and no symmetry.

Step 1: Find the y-intercept

To find the y-intercept of the equation, set \( x = 0 \) and solve for \( y \). The equation becomes \( y = 0^2 - 2 imes 0 + 1 \), simplifying to \( y = 1 \). So, the y-intercept is at the point \( (0, 1) \).

Step 2: Find the x-intercepts

To find the x-intercepts, set \( y = 0 \) and solve for \( x \). The equation is \( 0 = x^2 - 2x + 1 \). This can be factored as \( (x - 1)^2 = 0 \), giving \( x = 1 \). Therefore, the x-intercept is at \( (1, 0) \).

Step 3: Determine symmetry

To check for symmetry, examine the equation with respect to the y-axis, x-axis, and origin.1. **Y-axis symmetry**: Substitute \( -x \) for \( x \). If \( y = (-x)^2 - 2(-x) + 1 \) simplifies to the original equation, it has y-axis symmetry. - It simplifies to \( y = x^2 + 2x + 1 \), which is not the same, so no y-axis symmetry.2. **X-axis symmetry**: Substitute \( -y \) for \( y \). If \( -y = x^2 - 2x + 1 \) simplifies to the original equation, it has x-axis symmetry. - It doesn't, so no x-axis symmetry.3. **Origin symmetry**: Substitute \( -x \) and \( -y \) into the equation. If \( -y = (-x)^2 - 2(-x) + 1 \) simplifies to the original equation, it has origin symmetry. - It simplifies to \( -y = x^2 + 2x + 1 \), which is not the same, so no origin symmetry.

Step 4: Sketch the graph

Now that we have the intercepts and know there is no symmetry, plot the points \( (0, 1) \) for the y-intercept and \( (1, 0) \) for the x-intercept on a coordinate plane. The equation \( y = x^2 - 2x + 1 \) is a parabola that opens upwards. The vertex can be found by completing the square or using the vertex formula \( x = -\frac{b}{2a} = 1 \), which means the vertex is at \( (1, 0) \). However, we already have that point as the x-intercept. Draw a smooth curve through these points to plot the parabola.

Key Concepts

Finding Intercepts

When graphing a quadratic equation like \[ y = x^2 - 2x + 1 \]intercepts are crucial points where the graph crosses the axes.

• Y-intercept: To find the y-intercept, set \( x = 0 \). Plugging this into the equation gives:\[ y = (0)^2 - 2(0) + 1 = 1 \] Thus, the graph crosses the y-axis at the point \( (0, 1) \).


• X-intercepts: These are found by setting \( y = 0 \). For this equation, it involves solving\[ 0 = x^2 - 2x + 1 \].Factoring, we get:\[ (x - 1)^2 = 0 \]which simplifies to \( x = 1 \).
  This means the graph touches the x-axis at the point \( (1, 0) \).

Knowing the intercepts helps in confidently sketching your graph, as these are definitive anchor points.

Symmetry in Graphs

Understanding symmetry in graphs can reveal a lot about the shape and behavior of the equation. A graph is symmetric if it looks the same after a particular transformation, like flipping or rotating.

• Y-axis Symmetry: Substitute \( -x \) for \( x \) in the equation. Examine if it transforms back to the original equation. Here, substituting gives:\[ y = (-x)^2 - 2(-x) + 1 = x^2 + 2x + 1 \],which is different, indicating no y-axis symmetry.


• X-axis Symmetry: Substitute \( -y \) for \( y \). It would require the equation\[ -y = x^2 - 2x + 1 \]to match the original, but this isn't the case, proving no x-axis symmetry.


• Origin Symmetry: Replace both \( x \) with \( -x \) and \( y \) with \( -y \). If the equation remains unchanged, it shows symmetry around the origin.For this equation, \[ -y = (-x)^2 - 2(-x) + 1 = x^2 + 2x + 1 \],demonstrates no origin symmetry.

Knowing symmetry helps in both graphing and simplifying analysis of the parabola.

Vertex of a Parabola

In a quadratic equation, the vertex is the parabola's peak or valley. It's the point where the graph changes direction. Finding the vertex is straightforward with the formula:\[ x = -\frac{b}{2a} \]for an equation in the form \( ax^2 + bx + c \).
For \[ y = x^2 - 2x + 1 \], \( a = 1 \), \( b = -2 \), and \( c = 1 \). Plug these into the vertex formula to find:\[ x = -\frac{-2}{2 \times 1} = 1 \].Substitute back into the equation to find \( y \):\[ y = (1)^2 - 2(1) + 1 = 0 \].Hence, the vertex is at \( (1, 0) \),which for this equation, also happens to be the x-intercept. The vertex serves as the focal point for the graph, showing where the parabola is tightest before it opens up.