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Chapter 8: Problem 61

Find the \(x\) -and \(y\) -intercepts of the graph of the quadratic function. See Example \(9 .\) $$ f(x)=-2 x^{2}+4 x $$

Short Answer

Expert verified
The x-intercepts are at \((0,0)\) and \((2,0)\); the y-intercept is at \((0,0)\).

Step by step solution

01

Understanding Intercepts

The intercepts of a graph are the points where the graph crosses the axes. For the quadratic function \( f(x) = -2x^2 + 4x \), we will find the \(x\)-intercepts and \(y\)-intercepts separately.
02

Finding the y-intercept

The \(y\)-intercept is where the graph of the function crosses the \(y\)-axis. This happens when \(x = 0\). Substitute \(x = 0\) into the function: \[f(0) = -2(0)^2 + 4(0) = 0\]. Thus, the \(y\)-intercept is at the point \((0, 0)\).
03

Finding the x-intercepts

The \(x\)-intercepts occur where the function equals zero \(f(x) = 0\). Solve the equation:\[-2x^2 + 4x = 0\]Factor out \(-2x\):\[-2x(x - 2) = 0\] Set each factor to zero: \(-2x = 0\) leads to \(x = 0\), and \(x - 2 = 0\) leads to \(x = 2\).Thus, the \(x\)-intercepts are at the points \((0, 0)\) and \((2, 0)\).
04

Confirming the Intercepts

For both the \(x\)-intercepts and \(y\)-intercepts, the calculations confirm that the graph intersects at \((0, 0)\) and \((2, 0)\). These points satisfy the original equation \(f(x) = -2x^2 + 4x\) being zero or evaluated at \(y\).

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Key Concepts

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

Understanding X-Intercepts
The x-intercepts of a quadratic function are crucial points where the graph crosses the x-axis. This happens when the function's output, or y-value, is zero. For the quadratic function given by \( f(x) = -2x^2 + 4x \), finding the x-intercepts involves setting the equation equal to zero:
  • Start with the equation \(-2x^2 + 4x = 0\).
  • Factor out common terms to simplify: \(-2x (x - 2) = 0\).
Setting each factor equals to zero yields two solutions:
  • \(-2x = 0\) leading to \(x = 0\).
  • \(x - 2 = 0\) leading to \(x = 2\).
Thus, the x-intercepts are at points \((0,0)\) and \((2,0)\). These are the points where the graph touches or crosses the x-axis.
Understanding Y-Intercepts
The y-intercept of a quadratic function is another special point that shows where the graph crosses the y-axis. This crossing happens when the input value, \(x\), is zero. You can find the y-intercept of \( f(x) = -2x^2 + 4x \) by substituting \(x = 0\) into the function:
  • Plug \(x = 0\) into \(-2(0)^2 + 4(0) = 0\).
So, the y-intercept is at the point \((0, 0)\). This point confirms that the graph passes through the origin. It's not just a random point, but rather a key feature that helps in understanding the graph's overall shape. The y-intercept tells you immediately that when you start plotting from the origin, this will be a key anchor for drawing the parabola correctly.
Graphing Quadratics
Graphing a quadratic function like \( f(x) = -2x^2 + 4x \) involves plotting its key features, starting with the intercepts. Here's how you do it:
  • First, mark the intercepts on the graph. The x-intercepts are at \((0, 0)\) and \((2, 0)\), while the y-intercept is also at \((0, 0)\).
  • Next, observe the shape of the graph. Since the leading coefficient \(-2\) is negative, the parabola opens downward.
  • Identify the vertex as it often is either at the midpoint of the x-intercepts if they exist, or where the function changes direction.
To ensure accuracy, calculate a few more points between and beyond the intercepts by selecting x-values and solving the function. This provides a clearer picture of the curve's nature. When graphed, this quadratic reveals its symmetrical and downward-opening nature, with its axis of symmetry along x = 1, halfway between the intercepts. Visualizing your function this way helps in understanding how changes to the equation affect the graph's structure.

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