/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Let \((h, k)\) be the coordinate... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 23

Let \((h, k)\) be the coordinates of the vertex of a parabola. Then \(h\) equals the average of the two real zeros of the function (if they exist). For each of the following use this to find \(h,\) and then put the equations into the vertex form, \(y=a(x-h)^{2}+k\) a. A parabola with equation \(y=x^{2}+2 x-8\) b. A parabola with equation \(y=-x^{2}-3 x+4\)

Short Answer

Expert verified
a. \ h = -1, \ y = (x+1)^2 - 9\ b. \ h = -1.5, \ y = -(x+1.5)^2 + 6.25\

Step by step solution

01

Find the Zeros of the Parabola (Part a)

To find the zeros of the parabola with the equation \(y = x^{2} + 2x - 8\), solve for \(x\) when \(y = 0\):\[x^{2} + 2x - 8 = 0\]Use the quadratic formula, \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \, where \(a = 1 \, \ b = 2 \, \ c = -8\):\[x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)}\x = \frac{-2 \pm \sqrt{4 + 32}}{2}\x = \frac{-2 \pm \sqrt{36}}{2}\x = \frac{-2 \pm 6}{2}\]The solutions are \ x = 2 \ and \ x = -4\.
02

Calculate h (Part a)

The value of \(h\) is the average of the zeros: \[h = \frac{2 + (-4)}{2} = \frac{-2}{2} = -1\]
03

Find k (Part a)

Use the vertex form equation. Since the vertex \((-h, k)\) is given by h = -1, substitute \ x = -1\ into the original equation \ y = x^{2} + 2x - 8\ to find k: \[k = (-1)^{2} + 2(-1) - 8\k = 1 - 2 - 8 = -9\]
04

Write the Vertex Form Equation (Part a)

Now, substitute \(h\) and \(k\) back into the vertex form equation \ y = a(x-h)^2 + k\: \[y = 1(x+1)^{2} - 9\]
05

Find the Zeros of the Parabola (Part b)

For the parabola with the equation \(y = -x^{2} - 3x + 4\), solve for \(x\) when \(y = 0\): \[-x^{2} - 3x + 4 = 0\]Use the quadratic formula: \[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-1)(4)}}{2(-1)}\x = \frac{3 \pm \sqrt{9 + 16}}{-2}\x = \frac{3 \pm \sqrt{25}}{-2}\x = \frac{3 \pm 5}{-2}\]The solutions are \ x = -4\ and \ x = 1\.
06

Calculate h (Part b)

The value of \(h\) is the average of the zeros: \[h = \frac{-4 + 1}{2} = -\frac{3}{2} = -1.5\]
07

Find k (Part b)

Since \(h = -1.5,\), substitute \ x = -1.5\ into the original equation \y = -x^{2} - 3x + 4\ to find k: \[k = -(-1.5)^{2} - 3(-1.5) + 4\k = -2.25 + 4.5 + 4 = 6.25 \]
08

Write the Vertex Form Equation (Part b)

Now, substitute \(h\) and \(k\) back into the vertex form equation \ y = a(x-h)^2 + k\: \[y = -1(x + 1.5)^{2} + 6.25\]

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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 fundamental tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula itself is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, the coefficients \(a\), \(b\), and \(c\) come from the standard form of the quadratic equation.
This formula provides us with two solutions, known as the roots or zeros of the quadratic function. These solutions can help us transform the equation into its vertex form and find important features like the axis of symmetry and the vertex coordinates.
Zeros of a Function
The zeros of a function are the \(x\)-values where the function equals zero. In other words, they are the solutions to the equation \(f(x) = 0\). To find the zeros of a quadratic function, you can use the quadratic formula:
  • Set the quadratic equation to zero: \(ax^2 + bx + c = 0\).
  • Use the quadratic formula to solve for \(x\).
  • The solutions you find are the zeros.

For example, for the equation \(y = x^2 + 2x - 8\), we set \(y\) to 0 and solve using the quadratic formula: \(x = \frac{-2 \pm \sqrt{36}}{2}\), resulting in \(x = 2 \) and \(x = -4\). Thus, the zeros are 2 and -4.
Vertex Coordinates
The vertex of a parabola is its highest or lowest point, depending on the parabola's orientation. It is represented as \((h, k)\) where \(h\) is the x-coordinate and \(k\) is the y-coordinate. To find the vertex:
  • Find the zeros of the function, if they exist.
  • Calculate \(h\), the average of the zeros: \(h = \frac{x_1 + x_2}{2}\).
  • Plug \(h\) back into the original equation to find \(k\).

For example, for \(y = x^2 + 2x - 8\), the zeros are 2 and -4. So, \(h\) is \( \frac{2 + (-4)}{2} = -1 \). Then, \(k\) is found by plugging \(x = -1\) into the original equation, which results in \(k = -9\). Thus, the vertex is \((-1, -9)\).
Standard Form to Vertex Form
Converting a quadratic equation from its standard form, \(y = ax^2 + bx + c\), to its vertex form, \(y = a(x-h)^2 + k\), involves several steps:
  • First, find the zeros using the quadratic formula.
  • Find \(h\) by averaging the zeros.
  • Find \(k\) by substituting \(h\) into the original equation.
  • Rewrite the equation using \(h\) and \(k\) in the vertex form formula.

For instance, let's convert \(y = x^2 + 2x - 8\) to vertex form:
  • Find the zeros: \(2\) and \(-4\).
  • Calculate \(h\): \(h = -1\).
  • Calculate \(k\): \(k = -9\).
  • Substitute into vertex form formula: \(y = 1(x + 1)^2 - 9\).

The vertex form makes it easier to graph the parabola and understand its geometry.

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Most popular questions from this chapter

The wind chill temperature is the apparent temperature caused by the extra cooling from the wind. A rule of thumb for estimating the wind chill temperature for an actual temperature \(t\) that is above \(0^{\circ}\) Fahrenheit is \(W(t)=t-1.5 S_{0}\), where \(S_{0}\) is any given wind speed in miles per hour. a. If the wind speed is 25 mph and the actual temperature is \(10^{\circ} \mathrm{F}\), what is the wind chill temperature? We know how to convert Celsius to Fahrenheit; that is, we can write \(t=F(x),\) where \(F(x)=32+\frac{9}{5} x,\) with \(x\) the number of degrees Celsius and \(F(x)\) the equivalent in degrees Fahrenheit. b. Construct a function that will give the wind chill temperature as a function of degrees Celsius. c. If the wind speed is \(40 \mathrm{mph}\) and the actual temperature is \(-10^{\circ} \mathrm{C},\) what is the wind chill temperature?

Let \((h, k)\) be the coordinates of the vertex of a parabola. Then \(h\) is equal to the average of the two real zeros of the function (if they exist). For parts (a) and (b) use this to find \(h\), and then construct an equation in vertex form, \(y=a(x-h)^{2}+k\). a. A parabola with \(x\) -intercepts of 4 and 8 , and a \(y\) -intercept of 32 b. A parabola with \(x\) -intercepts of -3 and \(1,\) and a \(y\) -intercept of -1 c. Can you find the equation of a parabola knowing only its \(x\) -intercepts? Explain.

For the following quadratic functions in vertex form, \(f(x)=a(x-h)^{2}+k,\) determine the values for \(a, h,\) and \(k\) Then compare each to \(f(x)=x^{2},\) and identify which constants represent a stretch/compression factor, or a shift in a particular direction. a. \(p(x)=5(x-4)^{2}-2\) b. \(g(x)=\frac{1}{3}(x+5)^{2}+4\) c. \(h(x)=-0.25\left(x-\frac{1}{2}\right)^{2}+6\) d. \(k(x)=-3(x+4)^{2}-3\)

(Graphing program required.) At low speeds an automobile engine is not at its peak efficiency; efficiency initially rises with speed and then declines at higher speeds. When efficiency is at its maximum, the consumption rate of gas (measured in gallons per hour) is at a minimum. The gas consumption rate of a particular car can be modeled by the following equation, where \(G\) is the gas consumption rate in gallons per hour and \(M\) is speed in miles per hour: \(G=0.0002 M^{2}-0.013 M+1.07\) a. Construct a graph of gas consumption rate versus speed. Estimate the minimum gas consumption rate from your graph and the speed at which it occurs. b. Using the equation for \(G,\) calculate the speed at which the gas consumption rate is at its minimum. What is the minimum gas consumption rate? c. If you travel for 2 hours at peak efficiency, how much gas will you use and how far will you go? d. If you travel at \(60 \mathrm{mph}\), what is your gas consumption rate? How long does it take to go the same distance that you calculated in part (c)? (Recall that travel distance = speed \(\times\) time traveled.) How much gas is required for the trip? e. Compare the answers for parts (c) and (d), which tell you how much gas is used for the same-length trip at two different speeds. Is gas actually saved for the trip by traveling at the speed that gives the minimum gas consumption rate? f. Using the function \(G,\) generate data for gas consumption rate measured in gallons per mile by completing the following table. Plot gallons per mile (on the vertical axis) vs. miles per hour (on the horizontal axis). At what speed is gallons per mile at a minimum? g. Add a fourth column to the data table. This time compute miles/gal \(=\mathrm{mph} /(\mathrm{gal} / \mathrm{hr}) .\) Plot miles per gallon vs. miles per hour. At what speed is miles per gallon at a maximum? This is the inverse of the preceding question; we are normally used to maximizing miles per gallon instead of minimizing gallons per mile. Does your answer make sense in terms of what you found for parts (b) and (f)?

If we know the radius and depth of a parabolic reflector, we also know where the focus is. a. Find a generic formula for the focal length \(f\) of a parabolic reflector expressed in terms of its radius \(R\) and depth \(D\). The focal length \(\left|\frac{1}{4 a}\right|\) is the distance between the vertex and the focal point. Assume \(a>0\). b. Under what conditions does \(f=D ?\)
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