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Chapter 8: Problem 56

Find the maximum or minimum value of \(y\) for each function. $$y=6+5 x^{2}$$

Short Answer

Expert verified
The minimum value of \(y\) is 6.

Step by step solution

01

Identify the type of function

The given function is a quadratic function of the form \[y = ax^2 + bx + c\]. In this case, the function is \(y = 5x^2 + 6\). For quadratic functions, the maximum or minimum value of \(y\) occurs at the vertex.
02

Determine the direction of the parabola

Since the coefficient of \(x^2\) (which is 5) is positive, the parabola opens upwards. This means that the function has a minimum value.
03

Find the x-coordinate of the vertex

The x-coordinate of the vertex of a quadratic function \[y = ax^2 + bx + c\] is given by \[x = -\frac{b}{2a}\]. In this problem, \(a = 5\) and \(b = 0\). So the x-coordinate is: \[x = -\frac{0}{2 \times 5} = 0\].
04

Find the y-coordinate of the vertex

To find the y-coordinate of the vertex, substitute the x-coordinate of the vertex back into the function: \[y = 6 + 5(0)^2 = 6\].
05

Determine the minimum value

Since the vertex represents the minimum point of the parabola, the minimum value of \(y\) is \(y = 6\).

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Key Concepts

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

parabola
A parabolic curve, or simply a parabola, is the graph of a quadratic function. Parabolas can open upwards or downwards depending on the coefficient of the squared term (x虏). Here鈥檚 what to keep in mind:
  • If the coefficient (a) of x虏 is positive, the parabola opens upwards.
  • If the coefficient (a) of x虏 is negative, the parabola opens downwards.
Parabolas are symmetric about a vertical line called the axis of symmetry. The shape of the parabola makes them interesting in various real-life applications, from physics to engineering.
vertex
The vertex is a crucial point on a parabola. It represents the highest or lowest point on the curve, depending on the direction of the parabola. For the quadratic function \[y = ax^2 + bx + c\], the x-coordinate of the vertex can be found using the formula
\tag{x = -\frac{b}{2a}}.
Once you have the x-coordinate, substitute it back into the original function to find the y-coordinate. So, the vertex is at the point (x, y). This is particularly useful because:
  • It helps to determine the maximum or minimum value of the function.
  • It鈥檚 the point where the function changes direction.
minimum value
In a quadratic function where the parabola opens upward (a > 0), the vertex represents the minimum value of the function. Conversely, if the parabola opens downward (a < 0), the vertex represents the maximum value. Here鈥檚 a quick way to find both:
  • After finding the x-coordinate of the vertex using \[x = -\frac{b}{2a}\], substitute it back into the function to get the y-coordinate.
For the given exercise, we determined that the minimum value of y for the function \[y = 6 + 5x^2\] is 6.
quadratic formula
The quadratic formula is often used to find the roots of a quadratic equation \[ax^2 + bx + c = 0\]. Although it is not directly needed to find the vertex, it鈥檚 essential to understand it because:
  • It can help solve quadratic equations for x.
  • Understanding its derivation strengthens your grasp of quadratic functions.
The formula is given by:
\tag{x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}}.
Using this formula, you can find where the parabola intersects the x-axis. This is critical for both theoretical and practical applications.

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Most popular questions from this chapter

Find all real or imaginary solutions to each equation. Use the method of your choice. $$\left(p+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Find \(b^{2}-4 a c\) and the number of real solutions to each equation. $$-x^{2}+3 x-4=0$$

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Diagonal of a square. The diagonal of a square is \(2 \mathrm{m}\) longer than a side. Find the length of a side.

For each equation, find approximate solutions rounded to two decimal places. $$x^{2}-7.3 x+12.5=0$$

Solve each inequality. State the solution set using interval notation when possible. \(t^{2}<3(2 t-3)\)
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