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Chapter 2: Problem 7

Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=3 x^{2}-5 x+1\)

Short Answer

Expert verified
The vertex of the parabola is at the point \(\left(\frac{5}{6}, \frac{-1}{4}\right)\). There are two x-intercepts: \(\left(\frac{5 - \sqrt{13}}{6}, 0\right)\) and \(\left(\frac{5 + \sqrt{13}}{6}, 0\right)\). The parabola opens upwards. Plot the vertex and x-intercepts on the coordinate plane and sketch the parabola accordingly.

Step by step solution

01

Find the vertex

The vertex of a parabola is the point where the parabola changes direction and is found using the formula: \[h = \frac{-b}{2a}\] In this case, a = 3 and b = -5. Plugging these values into the formula, we get: \[h = \frac{-(-5)}{2(3)}\] \[h = \frac{5}{6}\] Now, we need to find the value of f(h) to get the vertex: \[f(h) = 3\left(\frac{5}{6}\right)^2 - 5\left(\frac{5}{6}\right) + 1\] After calculating, we get: \[f(h) = \frac{-1}{4}\] So, the vertex of the parabola is at the point \(\left(\frac{5}{6}, \frac{-1}{4}\right)\).
02

Find the x-intercepts (if any)

To find the x-intercepts, we need to set f(x) = 0 and solve for x. This can be done using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Plugging in the values a = 3, b = -5, and c = 1, we get: \[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(1)}}{2(3)}\] \[x = \frac{5 \pm \sqrt{25 - 12}}{6}\] \[x = \frac{5 \pm \sqrt{13}}{6}\] There are two x-intercepts, given by the points \(\left(\frac{5 - \sqrt{13}}{6}, 0\right)\) and \(\left(\frac{5 + \sqrt{13}}{6}, 0\right)\).
03

Sketch the parabola

To sketch the parabola, use the following points and properties: 1. Vertex: \(\left(\frac{5}{6}, \frac{-1}{4}\right)\) 2. x-intercepts: \(\left(\frac{5 - \sqrt{13}}{6}, 0\right)\) and \(\left(\frac{5 + \sqrt{13}}{6}, 0\right)\) 3. Since the coefficient of the x虏 term is positive (a = 3 > 0), the parabola opens upwards. Now, plot the vertex and x-intercepts on the coordinate plane and draw a parabola that goes through these points. It should open upwards and be symmetric with respect to the vertex's vertical line (x-axis at x = 5/6). This will give you a rough sketch of the parabola.

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Key Concepts

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

Vertex of a Parabola
Understanding the vertex of a parabola is fundamental when studying quadratic functions. The vertex is the peak or the lowest point on the parabola, depending on whether it opens upwards or downwards, respectively. To find the vertex, we use the formula: \[ h = \frac{-b}{2a} \]
To determine the y-coordinate of the vertex (k), we substitute x = h into the quadratic equation. In our case, with the quadratic function \(f(x)=3x^2-5x+1\), after using the formula we find the vertex to be \(\left(\frac{5}{6}, \frac{-1}{4}\right)\). The vertex splits the parabola into two mirror images, which is key for graphing and understanding the parabola's shape and direction.
X-intercepts of a Quadratic Function
The points where a parabola crosses the x-axis are known as the x-intercepts or roots. By setting the quadratic function equal to zero, \(f(x) = 0\), and solving for x, we can find its x-intercepts. The quadratic formula is the tool we often use for this: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our function \(f(x) = 3x^2 - 5x + 1\), applying the quadratic formula yields two x-intercepts at \(\left(\frac{5 - \sqrt{13}}{6}, 0\right)\) and \(\left(\frac{5 + \sqrt{13}}{6}, 0\right)\). These points are crucial for graphing the parabola as they mark the locations where the curve meets the x-axis.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using this formula, you will always be able to find the solutions, known as the roots, for any quadratic equation - regardless if they're real or complex numbers. For the quadratic function in our exercise, the quadratic formula helped us find the x-intercepts. Remember, 'b' stands for the coefficient of the x term, 'a' is the coefficient of the x虏 term, and 'c' is the constant. The symbol \(\pm\) indicates that you will solve for x twice: once using the positive square root to find one x-intercept, and once using the negative square root to find the other.

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Most popular questions from this chapter

In Exercises 1-18, find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=x^{2}+x-6\)

A ball is thrown straight upward from the ground and attains a height of \(s(t)=-16 t^{2}+128 t+\) \(4 \mathrm{ft}\) above the ground after \(t \mathrm{sec}\). When does the ball reach the maximum height? What is the maximum height?

According to a study, the outof-pocket cost to senior citizens for health care, \(f(t)\) (as a percentage of income), in year \(t\) where \(t=0\) corresponds to 1977 , is given by $$ f(t)=\left\\{\begin{array}{ll} \frac{2}{7} t+12 & \text { if } 0 \leq t \leq 7 \\ t+7 & \text { if } 7

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Patricia's neighbor, Juanita, also wishes to have a rectangular-shaped garden in her backyard. But Juanita wants her garden to have an area of \(250 \mathrm{ft}^{2}\). Letting \(x\) denote the width of the garden, find a function \(f\) in the variable \(x\) giving the length of the fencing required to construct the garden. What is the domain of the function? Hint: Refer to the figure for Exercise 26. The amount of fencing required is equal to the perimeter of the rectangle, which is twice the width plus twice the length of the rectangle.
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