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3 years ago

Quadratic graphs: y = x2 + 4x + 1 I need to work out the coordinates. Help.

Mathematics

2 answers:

Ray Of Light [21]3 years ago

4 0

If x =1,
y= 1^2 +4(1) +1
=7
x=0, y=1

Iteru [2.4K]3 years ago

3 0

Answer:

Basically what you have to do to draw the graph is creating a bunch of points that are part of the equation

to do so, you assign different values to x and find the corresponding y

for example

x=0 0^2 +4*0 +1 = 1 therefore y = 1

x=1 1^2+4*1 +1 = 6 therefore y= 6

and so on....

x=-1 y=-2

and as many as you need

this is valid...only if I have understood your question :-)

Step-by-step explanation:

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Find an example for each of vectors x, y ∈ V in R.

rjkz [21]

(a) Both conditions are satisfied with x = (1, 0) for $\mathbb R^2$ and x = (1, 0, 0) for $\mathbb R^3$ :

||(1, 0)|| = √(1² + 0²) = 1

max{1, 0} = 1

||(1, 0, 0)|| = √(1² + 0² + 0²) = 1

max{1, 0, 0} = 1

(b) This is the well-known triangle inequality. Equality holds if one of x or y is the zero vector, or if x = y. For example, in $\mathbb R^2$ , take x = (0, 0) and y = (1, 1). Then

||x + y|| = ||(0, 0) + (1, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2

||x|| + ||y|| = ||(0, 0)|| + ||(1, 1)|| = √(0² + 0²) + √(1² + 1²) = √2

The left side is strictly smaller if both vectors are non-zero and not equal. For example, if x = (1, 0) and y = (0, 1), then

||x + y|| = ||(1, 0) + (0, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2

||x|| + ||y|| = ||(1, 0)|| + ||(0, 1)|| = √(1² + 0²) + √(0² + 1²) = 2

and of course √2 < 2.

Similarly, in $\mathbb R^3$ you can use x = (0, 0, 0) and y = (1, 1, 1) for the equality, and x = (1, 0, 0) and y = (0, 1, 0) for the inequality.

x • y = ||x|| ||y|| cos(θ),

where θ is the angle between the vectors x and y. Both sides are scalar, so taking the norm gives

||x • y|| = ||(||x|| ||y|| cos(θ)|| = ||x|| ||y|| |cos(θ)|

Suppose x = (0, 0) and y = (1, 1). Then

||x • y|| = |(0, 0) • (1, 1)| = 0

||x|| • ||y|| = ||(0, 0)|| • ||(1, 1)|| = 0 • √2 = 0

For the inequality, recall that cos(θ) is bounded between -1 and 1, so 0 ≤ |cos(θ)| ≤ 1, with |cos(θ)| = 0 if x and y are perpendicular to one another, and |cos(θ)| = 1 if x and y are (anti-)parallel. You get everything in between for any acute angle θ. So take x = (1, 0) and y = (1, 1). Then

||x • y|| = |(1, 0) • (1, 1)| = |1| = 1

||x|| • ||y|| = ||(1, 0)|| • ||(1, 1)|| = 1 • √2 = √2

In $\mathbb R^3$ , you can use the vectors x = (1, 0, 0) and y = (1, 1, 1).

8 0

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VashaNatasha [74]

Answer:

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lawyer [7]

A. Helen is driving 65 mph
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Write the vector YZ in component form. Y(-9, 2) Z(3, -4)

Temka [501]

Answer:

$\displaystyle \overrightarrow{YZ} = \$

General Formulas and Concepts:
Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinate Planes

Coordinates (x, y)

Pre-Calculus

Vectors

Component Form: $\displaystyle \overrightarrow{XY} = \$

Step-by-step explanation:

Step 1: Define

Identify.

Point Y(-9, 2)

Point Z(3, -4)

Step 2: Find

Substitute in points [Vector Component Form]: $\displaystyle \overrightarrow{YZ} = \$
[Order of Operations] Evaluate: $\displaystyle \overrightarrow{YZ} = \$

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