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Chapter 2: Problem 43

\(f(x)=-2 x^2+8 x+7\)

Short Answer

Expert verified
The function \(f(x)=-2 x^2+8 x+7\) has the axis of symmetry at \(x = 2\), reaches its maximum value of 3 at \(x = 2\), and has x-intercepts at \(x = 3.5\) and \(x = -1\).

Step by step solution

01

Identify Coefficients

Recognize the general form of a quadratic function \(f(x) = ax^2 + bx + c\). In the function \(f(x)=-2 x^2+8 x+7\), \(a = -2\), \(b = 8\), and \(c = 7\).
02

Find the Axis of Symmetry

The formula of the axis of symmetry is \(x = -b / 2a\). Substituting \(a = -2\) and \(b = 8\), we get \(x = -8 / -4 = 2\).
03

Determine the Maximum or Minimum Value

Since \(a = -2\) is less than zero, the graph of the function opens downwards. Therefore, it has a maximum point. To find this point, substitute \(x = 2\) (from the axis of symmetry) into the function, we get \(f(2)=-2 (2)^2+8(2)+7=3\). So, the maximum point is (2, 3).
04

Find the X-Intercepts

The x-intercepts are found by setting \(f(x) = 0\), and solve for \(x\). This gives us \(-2x^2 + 8x + 7 = 0\). To solve this quadratic equation, we can use the quadratic formula \(x = [-b \pm sqrt(b^2 - 4ac)] / 2a\). After calculating, the roots are \(x = 3.5, -1\). So, the x-intercepts are (3.5, 0) and (-1, 0).

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Key Concepts

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

Axis of Symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For any quadratic function in the form of \(f(x) = ax^2 + bx + c\), the axis of symmetry can be found using the formula \(x = -\frac{b}{2a}\). This formula provides the x-coordinate that is the central point of the parabola.

In the given quadratic function \(f(x) = -2x^2 + 8x + 7\), the coefficients are \(a = -2\) and \(b = 8\). Plug this into the formula: \(x = -\frac{8}{2(-2)} = 2\).

This line of symmetry \(x = 2\) is where the maximum or minimum value of the quadratic function occurs on its graph.
Maximum Value
The maximum or minimum value of a quadratic function can be located using the vertex of the parabola. When the coefficient \(a\) is negative, the quadratic function will open downwards, meaning the parabola has a peak or maximum value.

To find this maximum, use the x-coordinate determined by the axis of symmetry. Substituting \(x = 2\) into the function \(f(x) = -2x^2 + 8x + 7\), we calculate the function's maximum point:

\(f(2) = -2(2)^2 + 8(2) + 7 = 3\).

Here, the maximum value is \(3\), occurring at the vertex \((2, 3)\). This point is critical as it represents the highest point on the graph.
X-Intercepts
X-intercepts are the points where the graph of a function crosses the x-axis. For a quadratic function, these are found by solving \(f(x) = 0\). The formula used to find these solutions is the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In our specific quadratic function \(-2x^2 + 8x + 7\), substitute the coefficients \(a = -2\), \(b = 8\), and \(c = 7\):

\(x = \frac{-8 \pm \sqrt{8^2 - 4(-2)(7)}}{2(-2)}\).

Solving it gives us the x-intercepts \((3.5, 0)\) and \((-1, 0)\).

These points denote where the parabola meets the x-axis, essential for understanding the graph's relation to the x-axis.

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Most popular questions from this chapter

MAKING AN ARGUMENT The point (1, 5) lies on the graph of a quadratic function with axis of symmetry x = ?1. Your friend says the vertex could be the point (0, 5). Is your friend correct? Explain.

\(y=\frac{3}{2} x^2-3 x+6\)

The path of a basketball thrown at an angle of \(45^{\circ} \mathrm{can}\) be modeled by \(y=-0.02 x^2+x+6\).

\(y=\frac{1}{2}(x-3)^2+2\)

The path of a shot put released at an angle of \(35^{\circ} \mathrm{can}\) be modeled by \(y=-0.01 x^2+0.7 x+6\).
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