Help plz don't know
1 answer:
You might be interested in
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:
Step-by-step explanation:
Prove: That the sum of the squares of 4 consecutive integers is an even integer.
An integer is a any directed number that has no decimal part or indivisible fractional part. Examples are: 4, 100, 0, -20,-100 etc.
Selecting 4 consecutive positive integers: 5, 6, 7, 8. Then;
= 25
= 36
= 49
= 64
The sum of the squares = 25 + 36 + 49 + 64
= 174
Also,
Selecting 4 consecutive negative integers: -10, -11, -12, -13. Then;
= 100
= 121
= 144
= 169
The sum of the squares = 100 + 121 + 144 + 169
= 534
Therefore, the sum of the squares of 4 consecutive integers is an even integer.