Problem 79 - Elementary Algebra

Question

Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist. $$ y=4 x 2-4 x+1 $$

Step-by-Step Solution

Answer

Vertex: \((\frac{1}{2}, 0)\); Y-intercept: \((0, 1)\); X-intercept: \((\frac{1}{2}, 0)\).

Step 1: Verify Correct Equation Format

Ensure the equation is written correctly. The provided equation is unclear. Assuming it is meant to be \( y = 4x^2 - 4x + 1 \).

Step 2: Identify the Vertex

The equation \( y = 4x^2 - 4x + 1 \) is in standard form \( y = ax^2 + bx + c \). Use the vertex formula \( x = -\frac{b}{2a} \). Here, \( a = 4 \), \( b = -4 \), \( c = 1 \). Calculate \( x = -\frac{-4}{2 \times 4} = \frac{1}{2} \).Substitute \( x = \frac{1}{2} \) into the equation to find \( y \):\[y = 4 \left( \frac{1}{2} \right)^2 - 4 \left( \frac{1}{2} \right) + 1 = 1 - 2 + 1 = 0.\]Thus, the vertex is at \( \left( \frac{1}{2}, 0 \right) \).

Step 3: Find the Y-Intercept

The y-intercept of a graph is where the graph intersects the y-axis, which is when \( x = 0 \). Substitute \( x = 0 \) into the equation:\[ y = 4(0)^2 - 4(0) + 1 = 1. \]Thus, the y-intercept is \( (0, 1) \).

Step 4: Identify the X-Intercepts

The x-intercepts occur where the graph intersects the x-axis, so set \( y = 0 \) and solve the equation:\[ 4x^2 - 4x + 1 = 0. \]This quadratic equation can be solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -4 \), \( c = 1 \):\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 4 \times 1}}{2 \times 4} = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4 \pm 0}{8}. \]The solution is \( x = \frac{1}{2} \).The x-intercept is \( (\frac{1}{2}, 0) \). There is only one x-intercept because the vertex touches the x-axis.

Key Concepts

Graphing Quadratic Functions

Graphing a quadratic function provides a visual representation of the solutions or roots of the equation. A quadratic function typically comes in the form \( y = ax^2 + bx + c \). This creates a parabolic curve, commonly known as a parabola.

When plotting these functions, you want to:

• Identify the direction of the parabola. The sign of \( a \) determines this: if \( a \) is positive, the parabola opens upwards (like a smile); if \( a \) is negative, it opens downwards (like a frown).
• Locate the vertex, which is the highest or lowest point on the graph, depending on the parabola's direction.
• Find the y-intercept, which is the point where the parabola crosses the y-axis. For a quadratic equation in standard form, it is simply the value of \( c \).

Understanding the symmetry in a parabola is essential. The parabola's axis of symmetry is a vertical line passing through the vertex, dividing the parabola into two mirror-image halves. This axis helps in predicting additional points on the graph.

Vertex of a Parabola

The vertex of a parabola is a crucial feature because it tells you the turning point. Calculating the vertex involves using the formula for \( x \), \( x = -\frac{b}{2a} \), from the quadratic equation \( y = ax^2 + bx + c \).

Here's how you find it:

• First, calculate the \( x \)-coordinate of the vertex using the formula.
• Then substitute this \( x \)-value back into the original quadratic equation to find the corresponding \( y \)-coordinate.
• Finally, combine these to form the vertex point \( (x, y) \).

In practice, for the equation \( y = 4x^2 - 4x + 1 \), you calculate \( x = -\frac{-4}{2 \times 4} = \frac{1}{2} \), then substitute back to find \( y \), resulting in the vertex \( \left( \frac{1}{2}, 0 \right) \). This indicates that the vertex is exactly at the same level on the y-axis as the x-intercepts.

X-Intercepts and Y-Intercepts

Intersections of the graph with the axes provide key insights into the behavior of the function. The intercepts allow you to find exact points where the graph touches the x-axis and y-axis.

Y-Intercept:

• Occurs where the graph crosses the y-axis. Easily found by setting \( x = 0 \) in the equation.
• For \( y = 4x^2 - 4x + 1 \), substituting \( x = 0 \) gives the y-intercept as \( (0, 1) \).

X-Intercepts:

• These points occur where the graph crosses the x-axis, requiring you to set \( y = 0 \) and solve for \( x \).
• Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for the x-intercepts.
• In many cases, as with \( y = 4x^2 - 4x + 1 \), the vertex itself is the only intercept, \( \left( \frac{1}{2}, 0 \right) \), when the discriminant \( b^2 - 4ac \) equals zero, signifying a perfect square trinomial.

These insights guide how you comprehend the structure and solutions of quadratic equations, linking algebraic expressions directly with their geometric form.