Problem 79 Graph. Find the vertex and the y... [FREE SOLUTION]

Chapter 9: Problem 79

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

Short Answer

Expert verified

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

Step by step solution

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 $.

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) $.

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) $.

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.

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Key Concepts

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

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.

91影视

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

Short Answer

Step by step solution

Verify Correct Equation Format

Identify the Vertex

Find the Y-Intercept

Identify the X-Intercepts

Key Concepts

Graphing Quadratic Functions

Vertex of a Parabola

X-Intercepts and Y-Intercepts

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