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Chapter 2: Problem 38

In Exercises \(35-44,\) graph the quadratic function. $$f(x)=-x^{2}+3 x+4$$

Short Answer

Expert verified
The graph has a downward opening parabola with vertex \( \left(\frac{3}{2}, \frac{7}{4}\right) \) and intercepts at (0, 4), (4, 0), and (-1, 0).

Step by step solution

01

Identify the Form of the Quadratic

The quadratic function given is \( f(x) = -x^2 + 3x + 4 \). It is in the standard form \( ax^2 + bx + c \), where \( a = -1 \), \( b = 3 \), and \( c = 4 \). Since \( a < 0 \), this parabola opens downward.
02

Find the Vertex

The vertex formula for a quadratic in the form \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Substituting the values, we get:\[x = -\frac{3}{2(-1)} = \frac{3}{2}\]To find the \( y \)-coordinate of the vertex, substitute \( x = \frac{3}{2} \) back into \( f(x) \):\[f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 4 = -\frac{9}{4} + \frac{9}{2} + 4 = \frac{7}{4}\]Thus, the vertex is \( \left(\frac{3}{2}, \frac{7}{4}\right) \).
03

Find the Y-Intercept

The \( y \)-intercept is found by setting \( x = 0 \) in the function:\[f(0) = -0^2 + 3 \cdot 0 + 4 = 4\]Thus, the \( y \)-intercept is \( (0, 4) \).
04

Determine the X-Intercepts

To find the \( x \)-intercepts, solve \( -x^2 + 3x + 4 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[x = \frac{-3 \pm \sqrt{3^2 - 4(-1)(4)}}{2(-1)} = \frac{-3 \pm \sqrt{9 + 16}}{-2} = \frac{-3 \pm 5}{-2}\]This gives us two solutions: \( x = 4 \) and \( x = -1 \). Therefore, the \( x \)-intercepts are \( (4, 0) \) and \( (-1, 0) \).
05

Sketch the Graph

To graph the quadratic function, plot the vertex \( \left(\frac{3}{2}, \frac{7}{4}\right) \), the \( y \)-intercept \( (0, 4) \), and the \( x \)-intercepts \( (4, 0) \) and \( (-1, 0) \). Since the parabola opens downward, draw a symmetric curve passing through these points. Ensure the axis of symmetry, \( x = \frac{3}{2} \), divides the parabola into two equal halves.

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Key Concepts

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

Vertex of a Parabola
A vertex is a significant point on a parabola where it changes direction. For the function \( f(x) = -x^2 + 3x + 4 \), the vertex represents the highest point of the parabola since it opens downward. The vertex can be found using the formula \( x = -\frac{b}{2a} \). When calculated for our function, it results in \( x = \frac{3}{2} \). To find the y-coordinate, substitute back into the original equation: \[ f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 4 = \frac{7}{4} \] Thus, the vertex is \( \left(\frac{3}{2}, \frac{7}{4}\right) \). This vertex divides the parabola into two equal sides.
X-Intercepts
The x-intercepts of a quadratic function are where the graph meets the x-axis. For these points, the y value is zero. To find the x-intercepts for \( f(x) = -x^2 + 3x + 4 \), set the equation to zero: \[ -x^2 + 3x + 4 = 0 \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve it: \[ x = \frac{-3 \pm \sqrt{9 + 16}}{-2} = \frac{-3 \pm 5}{-2} \] The solutions are \( x = 4 \) and \( x = -1 \). Therefore, the x-intercepts are at these points:
  • (4, 0)
  • (-1, 0)
Y-Intercept
The y-intercept occurs where the parabola crosses the y-axis, meaning the x value is zero. For \( f(x) = -x^2 + 3x + 4 \), substitute \( x = 0 \) to find the y-intercept: \[ f(0) = -0^2 + 3(0) + 4 = 4 \] Thus, the y-intercept for this quadratic function is at the point (0, 4). The y-intercept is a simple yet essential part of graphing as it provides a specific point where the parabola meets the vertical axis.
Graphing Parabolas
Graphing a parabola involves plotting key points and drawing a smooth curve through them. For our function \( f(x) = -x^2 + 3x + 4 \), start by plotting the vertex \( \left(\frac{3}{2}, \frac{7}{4}\right) \), which is the peak point since the parabola opens downward.
  • The y-intercept at (0, 4) helps guide the starting point of the parabola on the y-axis.
  • Plot the x-intercepts at (4, 0) and (-1, 0), which show where the curve crosses the x-axis.
  • Use symmetry about the line \( x = \frac{3}{2} \) to ensure both sides of the parabola mirror each other.
Finally, draw a smooth curve connecting these points, ensuring it opens downward as indicated by the negative coefficient of \( x^2 \). This visual model helps comprehend the nature and shape of the quadratic function efficiently.

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