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Chapter 2: Problem 5

Find the \(x\) -intercepts of the given function. $$y=4 x-4 x^{2}-1$$

Short Answer

Expert verified
The x-intercept is \(x = \frac{1}{2}\).

Step by step solution

01

Set the Function Equal to Zero

To find the x-intercepts, we need to set the function equal to zero since at the x-intercepts, the value of y is zero. So, we start with the equation:\[4x - 4x^2 - 1 = 0\]
02

Rearrange the Equation

Rearrange the equation to organize it into a standard quadratic form, which is \(ax^2 + bx + c = 0\). This gives us:\[-4x^2 + 4x - 1 = 0\]
03

Apply the Quadratic Formula

To solve for \(x\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = -4\), \(b = 4\), and \(c = -1\). Substitute these values into the formula:\[x = \frac{-4 \pm \sqrt{4^2 - 4(-4)(-1)}}{2(-4)}\]
04

Simplify the Expression

Simplify the expression under the square root first and then the entire expression:\[x = \frac{-4 \pm \sqrt{16 - 16}}{-8}\]\[x = \frac{-4 \pm \sqrt{0}}{-8}\]\[x = \frac{-4}{-8}\]\[x = \frac{1}{2}\]
05

Conclusion

Since the discriminant is zero and the simplified form gives us a single solution, the only x-intercept is at:\(x = \frac{1}{2}\).

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Key Concepts

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

quadratic formula
The quadratic formula is a powerful tool used to find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is stated as \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. These roots are the x-values where the graph of the quadratic function intersects the x-axis.
To use the quadratic formula:
  • Identify the coefficients \(a\), \(b\), and \(c\) in the quadratic equation.
  • Substitute these values into the formula.
  • Simplify the expression under the square root, called the discriminant, \(b^2 - 4ac\).
  • Continue simplifying to find the values of \(x\).
Simplifying can involve working through complex numbers if the discriminant is negative.
Thankfully, our given exercise has a discriminant of zero, resulting in a straightforward single solution.
solving quadratic equations
Solving quadratic equations involves finding the values of \(x\) where the equation \(ax^2 + bx + c = 0\) holds true. The solutions can be:
  • Real and distinct, if the discriminant (\(b^2 - 4ac\)) is positive.
  • Real and identical, if the discriminant is zero.
  • Complex, if the discriminant is negative.
In our problem, we followed these steps:
First, we set the function equal to zero:
\

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Most popular questions from this chapter

Each of the graphs of the functions has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. [Recall that if \(f(x)=a x^{2}+b x+c,\) then \(f(x)\) has a relative minimum point when \(a>0\) and a relative maximum point when \(a<0.1\) $$f(x)=1+6 x-x^{2}$$

Until recently hamburgers at the city sports arena cost \(\$ 4\) each. The food concessionaire sold an average of 10,000 hamburgers on a game night. When the price was raised to \(\$ 4.40,\) hamburger sales dropped oft to an average of 8000 per night. (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. (b) If the concessionaire has fixed costs of \(\$ 1000\) per night and the variable cost is \(\$ .60\) per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit.

Sketch the following curves, indicating all relative extreme points and inflection points. $$y=x^{4}-\frac{4}{3} x^{3}$$

Each of the graphs of the functions has one relative maximum and one relative minimum point. Find these points using the first-derivative test. Use a variation chart as in Example 1. $$f(x)=2 x^{3}+3 x^{2}-3$$

Match each observation (a)-(e) with a conclusion (A)-(E). Observations (a) The point \((3,4)\) is on the graph of \(f^{\prime}(x).\) (b) The point \((3,4)\) is on the graph of \(f(x).\) (c) The point \((3,4)\) is on the graph of \(f^{\prime \prime}(x).\) (d) The point \((3,0)\) is on the graph of \(f^{\prime}(x),\) and the point (3,4) is on the graph of \(f^{\prime \prime}(x).\) (e) The point \((3,0)\) is on the graph of \(f^{\prime}(x),\) and the point (3,-4) is on the graph of \(f^{\prime \prime}(x).\) Conclusions (A) \(f(x)\) has a relative minimum point at \(x=3.\) (B) When \(x=3,\) the graph of \(f(x)\) is concave up. (C) When \(x=3,\) the tangent line to the graph of \(y=f(x)\) has slope 4. (D) When \(x=3,\) the value of \(f(x)\) is 4. (E) \(f(x)\) has a relative maximum point at \(x=3.\)
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