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91Ó°ÊÓ

Write each of the following quadratic equations in function form (i.e., solve for \(y\) in terms of \(x\) ). Find the vertex and the \(y\) -intercept using any method. Finally, using these points, draw a rough sketch of the quadratic function. a. \(y+12=x(x+1)\) d. \(y-8 x=x^{2}+15\) b. \(2 x^{2}+6 x+14.4-2 y=0\) e. \(y+1=(x-2)(x+5)\) c. \(y+x^{2}-5 x=-6.25\) f. \(y+2 x(x-6)=20\)

Short Answer

Expert verified
Solved forms: (a) \(y = x^2 + x - 12\), (d) \(y = x^2 + 8x + 15\), (b) \(y = x^2 + 3x + 7.2\), (e) \(y = x^2 + 3x - 11\), (c) \(y = -x^2 + 5x - 6.25\), (f) \(y = 2x^2 - 12x - 20\). Vertices and \(y\)-intercepts given, graphs sketched.

Step by step solution

01

Title - Simplify to Function Form for (a)

Firstly, solve for \(y\) in terms of \(x\): Start with \(y + 12 = x(x + 1)\). Distribute on the right side to get \(y + 12 = x^2 + x\). Subtract 12 from both sides to isolate \(y\): \(y = x^2 + x - 12\).
02

Title - Find the Vertex for (a)

Using the completed square method or vertex form, note that for \(y = ax^2 + bx + c\), the vertex \(x\)-coordinate is \(x = -\frac{b}{2a}\). Here, \(a = 1\), \(b = 1\), so \(x = -\frac{1}{2} \cdot 1 = -0.5\). Substitute \(x = -0.5\) into \(y = x^2 + x - 12\) to find \(y\)-coordinate: \(y = (-0.5)^2 + (-0.5) - 12 = -12.25\). Thus, the vertex is \((-0.5, -12.25)\).
03

Title - Find the Y-Intercept for (a)

To find the \(y\)-intercept, set \(x = 0\) and solve for \(y\): \(y = 0^2 + 0 - 12 = -12\). Thus, the \(y\)-intercept is \((0, -12)\).
04

Title - Sketch the Graph for (a)

Using the vertex \((-0.5, -12.25)\) and the \(y\)-intercept \((0, -12)\), draw a rough sketch of the parabola. The parabola opens upwards because the coefficient of \(x^2\) is positive.
05

Title - Simplify to Function Form for (d)

Start with \(y - 8x = x^2 + 15\). Rearrange to solve for \(y\): \(y = x^2 + 8x + 15\)
06

Title - Find the Vertex for (d)

For the quadratic in the form \(y = x^2 + 8x + 15\), the vertex \(x\)-coordinate is \(x = -\frac{b}{2a} = -\frac{8}{2} = -4\). Substitute \(x = -4\) back into the equation to find \(y\): \(y = (-4)^2 + 8(-4) + 15 = -1\). Thus, the vertex is \((-4, -1)\).
07

Title - Find the Y-Intercept for (d)

To find the \(y\)-intercept, set \(x = 0\): \(y = 0^2 + 8(0) + 15 = 15\). So, the \(y\)-intercept is \((0, 15)\).
08

Title - Sketch the Graph for (d)

Using the vertex \((-4, -1)\) and the \(y\)-intercept \((0, 15)\), draw a rough sketch of the parabola. The parabola opens upwards.
09

Title - Simplify to Function Form for (b)

Start with \(2x^2 + 6x + 14.4 - 2y = 0\). Rearrange to solve for \(y\): \(2y = 2x^2 + 6x + 14.4\). Divide by 2: \(y = x^2 + 3x + 7.2\).
10

Title - Find the Vertex for (b)

The vertex \(x\)-coordinate is \(x = -\frac{b}{2a} = -\frac{3}{2}\). Substitute back in: \(y = (-1.5)^2 + 3(-1.5) + 7.2 = 4.95\). Thus, the vertex is \((-1.5, 4.95)\).
11

Title - Find the Y-Intercept for (b)

To find the \(y\)-intercept, set \(x = 0\): \(y = 0^2 + 3(0) + 7.2 = 7.2\). So, the \(y\)-intercept is \((0, 7.2)\).
12

Title - Sketch the Graph for (b)

Using the vertex \((-1.5, 4.95)\) and the \(y\)-intercept \((0, 7.2)\), draw a rough sketch of the parabola. The parabola opens upwards.
13

Title - Simplify to Function Form for (e)

Start with \(y + 1 = (x - 2)(x + 5)\). Expand the right side: \(y + 1 = x^2 + 3x - 10\). Subtract 1 from both sides: \(y = x^2 + 3x - 11\).
14

Title - Find the Vertex for (e)

The vertex \(x\)-coordinate is \(x = -\frac{b}{2a} = -\frac{3}{2} = -1.5\). Substitute back: \(y = (-1.5)^2 + 3(-1.5) - 11 = -13.25\). Thus, the vertex is \((-1.5, -13.25)\).
15

Title - Find the Y-Intercept for (e)

To find the \(y\)-intercept, set \(x = 0\): \(y = 0^2 + 3(0) - 11 = -11\). So, the \(y\)-intercept is \((0, -11)\).
16

Title - Sketch the Graph for (e)

Using the vertex \((-1.5, -13.25)\) and the \(y\)-intercept \((0, -11)\), draw a rough sketch of the parabola. The parabola opens upwards.
17

Title - Simplify to Function Form for (c)

Start with \(y + x^2 - 5x = -6.25\). Rearrange to solve for \(y\): \(y = -x^2 + 5x - 6.25\).
18

Title - Find the Vertex for (c)

The vertex \(x\)-coordinate is \(x = -\frac{b}{2a} = \frac{-5}{-2} = 2.5\). Substitute back: \(y = -(2.5)^2 + 5(2.5) - 6.25 = 0\). Thus, the vertex is \((2.5, 0)\).
19

Title - Find the Y-Intercept for (c)

To find the \(y\)-intercept, set \(x = 0\): \(y = -0^2 + 5(0) - 6.25 = -6.25\). So, the \(y\)-intercept is \((0, -6.25)\).
20

Title - Sketch the Graph for (c)

Using the vertex \((2.5, 0)\) and the \(y\)-intercept \((0, -6.25)\), draw a rough sketch of the parabola. The parabola opens downwards.
21

Title - Simplify to Function Form for (f)

Start with \(y + 2x(x - 6) = 20\). Expand and simplify: \(y + 2x^2 - 12x = 20\). Subtract 20 from both sides to isolate \(y\): \(y = 2x^2 - 12x - 20\).
22

Title - Find the Vertex for (f)

The vertex \(x\)-coordinate is \(x = -\frac{b}{2a} = \frac{12}{4} = 3\). Substitute back: \(y = 2(3)^2 - 12(3) - 20 = -38\). Thus, the vertex is \((3, -38)\).
23

Title - Find the Y-Intercept for (f)

To find the \(y\)-intercept, set \(x = 0\): \(y = 2(0)^2 - 12(0) - 20 = -20\). So, the \(y\)-intercept is \((0, -20)\).
24

Title - Sketch the Graph for (f)

Using the vertex \((3, -38)\) and the \(y\)-intercept \((0, -20)\), draw a rough sketch of the parabola. The parabola opens upwards.

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Key Concepts

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

Vertex Form
The vertex form of a quadratic equation is a helpful way to easily identify the vertex of the parabola. The general form of a quadratic equation is given by \(y = ax^2 + bx + c\). However, the vertex form rewrites it as \(y = a(x - h)^2 + k\), where \(h\) and \(k\) represent the coordinates of the vertex (\(h, k\)).

To convert a quadratic equation from standard form to vertex form, you can use the method of completing the square. Here's a simplified process:
  • Start with the quadratic equation in standard form \(y = ax^2 + bx + c\)
  • Factor out the coefficient \(a\) from the first two terms if \(a eq 1\)
  • Rewrite the quadratic and linear terms in perfect square form by completing the square
  • Add and subtract the same value inside the squared term
  • Simplify the equation to vertex form
Using this form makes it easy to identify the vertex and sketch the graph of the quadratic function. Vertex form tells us whether the parabola opens upwards or downwards (based on the sign of \(a\)) and where its highest or lowest point is located.
Y-Intercept
The y-intercept of a quadratic function is the point where the graph intersects the y-axis. To find this, simply set \(x = 0\) in the quadratic equation and solve for \(y\).

For example, consider the quadratic equation \(y = x^2 + x - 12\). By setting \(x = 0\):
\(y = 0^2 + 0 - 12 = -12\). Thus, the y-intercept is at the point \(0, -12\).

Finding the y-intercept gives you a key point that helps sketch the graph of the quadratic function. It represents the starting point where the parabola crosses the y-axis, making it a reliable anchor for visualizing the entire curve. Knowing the y-intercept, along with the vertex, allows for a more accurate sketch of the parabola.
Parabola
In coordinate geometry, a parabola is the graph of a quadratic function. It takes a distinct U-shape and can open either upwards or downwards.

Important characteristics of a parabola include:
  • **Vertex**: The highest or lowest point on the parabola, where it changes direction
  • **Axis of Symmetry**: A vertical line that runs through the vertex and divides the parabola into two mirror-image halves
  • **Direction of Opening**: Determined by the sign of the coefficient of \(x^2\) (positive signifies upwards, negative signifies downwards)
  • **Y-Intercept**: The point where the parabola crosses the y-axis
For instance, the quadratic function \(y = 2x^2 - 12x - 20\) describes an upward-opening parabola because its leading coefficient (2) is positive.

Understanding these elements helps in accurately sketching and analyzing quadratic functions. The vertex provides the key point from which the symmetry and shape of the parabola can be drawn, while the y-intercept offers an additional point of reference on the graph.

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Most popular questions from this chapter

A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle (see the accompanying figure). If the perimeter of the window is 20 feet (including the semicircle), what dimensions will admit the most light (maximize the area)? (Hint: Express \(L\) in terms of \(r\). Recall that the circumference of a circle \(=2 \pi r,\) and the area of a circle \(=\pi r^{2},\) where \(r\) is the radius of the circle.)

Students noticed that the path of water from a water fountain seemed to form a parabolic arc. They set a flat surface at the level of the water spout and measured the maximum height of the water from the flat surface as 8 inches and the distance from the spout to where the water hit the flat surface as 10 inches. Construct a function model for the stream of water.

A manufacturer sells children's wooden blocks packed tightly in a cubic tin box with a hinged lid. The blocks cost 3 cents a cubic inch to make. The box and lid material cost 1 cent per square inch. (Assume the sides of the box are so thin that their thickness can be ignored.) It costs 2 cents per linear inch to assemble the box seams. The hinges and clasp on the lid cost \(\$ 2.50,\) and the label costs 50 cents. a. If the edge length of the box is \(s\) inches, develop a formula for estimating the cost \(C(s)\) of making a box that's filled with blocks. b. Graph the function \(C(s)\) for a domain of 0 to \(20 .\) What section of the graph corresponds to what the manufacturer actually produces-boxes between 4 and 16 inches in edge length? c. What is the cost of this product if the cube's edge length is 8 inches? d. Using the graph of \(C(s)\), estimate the edge length of the cube when the total cost is \(\$ 100\)

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