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The base of a solid is the region in the first quadrant enclosed by the parabola y equals 4x2?
yes!
Is the graph of y equals 4x2-2x plus 5 a straight ine?
The equation y = 4x^2 + 5 is a parabola
How does y equals 4x2 plus 21x look in a graph?
It is a parabola with its vertex at the origin and the arms going upwards.
How do you factor 4x2 - y2?
4x2-y2 = (2x-y)(2x+y)
What is the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
How does the graph of y 4x2 compare to the graph of y x2?
Their noses are both at the origin, and they both open upward, but y=4x2 is a much skinnier parabola.
The base of a solid is the region in the first quadrant enclosed by the parabola y equals 4x2?
yes!
What are the roots of a parabola?
I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.
Is the graph of y equals 4x2-2x plus 5 a straight ine?
The equation y = 4x^2 + 5 is a parabola
How does y equals 4x2 plus 21x look in a graph?
It is a parabola with its vertex at the origin and the arms going upwards.
What is the focus of a parabola?
The focus of a parabola is a fixed point that lies on the axis of the parabola "p" units from the vertex. It can be found by the parabola equations in standard form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) depending on the shape of the parabola. The vertex is defined by (h,k). Solve for p and count that many units from the vertex in the direction away from the directrix. (your focus should be inside the curve of your parabola)
What is the equation of a prabola with the vertex 0 0 and focus 0 4?
x2 = 16y The standard formula for a parabola with its vertex at the origin (0, 0) and a given focus (and the y-axis as an axis of symmetry) is as follows: x2 = 4cy In this case, the c is the y value of the focus. The focus in this case was (0, 4), and the y value in the focus is 4. That makes the c = 4. Further, that makes the equation for this parabola x2 = 4 (c)y = 4 (4)y = 16y Given that the vertex was the origin, (0, 0), and the focus is (0, 4), we can conclude that the axis of symmetry is the y-axis because the y value of the focus is 0. We can also conclude that the parabola opens up, because the focus has a positive y value.
What is the equation of a parabola with vertex at 1 -3 and focus at 2 -3?
For a parabola with an axis of symmetry parallel to the x-axis, the equation of a parabola is given by: (y - k)² = 4p(x - h) Where the vertex is at (h, k), and the distance between the focus and the vertex is p (which can be calculated as p = x_focus - x_vertex). For the parabola with vertex (1, -3) and focus (2, -3) this gives: h = 1 k = -3 p = 2 - 1 = 1 → parabola is: (y - -3)² = 4×1(x - 1) → (y + 3)² = 4(x - 1) This can be expanded to: 4x = y² + 6y + 13 or x = (1/4)y² + (3/2)y + (13/4)
How do you factor 4x2 - y2?
4x2-y2 = (2x-y)(2x+y)
What is the point directly above the focus?
The point directly above the focus is the vertex of the parabola. The focus is a specific point on the axis of symmetry of the parabola, and the vertex is the point on the parabola that is closest to the focus.
What is the midpoint of the parabola between the focus and the directrix?
It is the apex of the parabola.
How do you write 4x2 plus y equals 16 using function notation?
First you need to solve for y. So write 4x2+y=16 so y=16-4x2 Now write f(x)=16-4x2
