graph PQR with vertices P(-2,3),Q (1,2), and R (3,-1) and its image after the translation ( x,y) -> ( x-2, y - 5)
1 answer:
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Answer:
the parabola can be written as:
f(x) = y = a*x^2 + b*x + c
first step.
find the vertex at:
x = -b/2a
the vertex will be the point (-b/2a, f(-b/2a))
now, if a is positive, then the arms of the parabola go up, if a is negative, the arms of the parabola go down.
The next step is to see if we have real roots by using the Bhaskara's equation:
Now, draw the vertex, after that draw the values of the roots in the x-axis, and now conect the points with the general draw of the parabola.
If you do not have any real roots, you can feed into the parabola some different values of x around the vertex
for example at:
x = (-b/2a) + 1 and x = (-b/2a) - 1
those two values should give the same value of y, and now you can connect the vertex with those two points.
If you want a more exact drawing, you can add more points (like x = (-b/2a) + 3 and x = (-b/2a) - 3) and connect them, as more points you add, the best sketch you will have.
Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
or
[ ,
,
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ ,
,
]
In conclusion, the two unit vectors are;
[ ,
,
]
and
[ ,
,
]
<em>Hope this helps!</em>