has gradient
which at the point (-1, 4, 3) has a value of
I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say , in which case we have
Then the derivative of at (-1, 4, 3) in the direction of
is
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.
