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The equation of the parabola ( y = 4x^2 ) can be rewritten in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex. Here, it is clear that the vertex is at the origin (0, 0) and ( 4p = 4 ), giving ( p = 1 ). The focus of the parabola is located at ( (h, k + p) ), so the focus is at the point ( (0, 1) ).
What else can I help you with?
What is the focus of the parabola y equals 4x2?
The equation ( y = 4x^2 ) represents a parabola that opens upwards. To find the focus, we can rewrite it in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Here, the vertex is at the origin ( (0, 0) ) and ( 4p = 4 ), so ( p = 1 ). Thus, the focus of the parabola is located at the point ( (0, 1) ).
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.
Which equation represents a parabola opening to the right with a vertex at the origin and a focus at (40)?
The equation that represents a parabola opening to the right with its vertex at the origin (0,0) and a focus at (4,0) is given by ( y^2 = 4px ), where ( p ) is the distance from the vertex to the focus. Since the focus is located at (4,0), ( p = 4 ). Therefore, the equation of the parabola is ( y^2 = 16x ).
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.
What is the focus of the parabola y equals 4x2?
The equation ( y = 4x^2 ) represents a parabola that opens upwards. To find the focus, we can rewrite it in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Here, the vertex is at the origin ( (0, 0) ) and ( 4p = 4 ), so ( p = 1 ). Thus, the focus of the parabola is located at the point ( (0, 1) ).
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)
Which equation represents a parabola opening to the right with a vertex at the origin and a focus at (40)?
The equation that represents a parabola opening to the right with its vertex at the origin (0,0) and a focus at (4,0) is given by ( y^2 = 4px ), where ( p ) is the distance from the vertex to the focus. Since the focus is located at (4,0), ( p = 4 ). Therefore, the equation of the parabola is ( y^2 = 16x ).
What is Focus (34) directrix y -2?
The focus of a parabola is a specific point that defines its shape, while the directrix is a line used in the definition of a parabola. If the directrix is given as ( y = -2 ), the parabola opens either upwards or downwards. The focus would be located at a point above or below this directrix, depending on the orientation of the parabola. Specifically, for a parabola that opens upwards, the focus would be positioned at ( (h, k + p) ), where ( p ) is the distance from the vertex to the focus, and the vertex would be located at ( (h, -2 + p) ).
What is the primary focal chord of a parabola?
The primary focal chord of a parabola is a line segment that passes through the focus of the parabola and has its endpoints on the parabola itself. For a standard parabola defined by the equation (y^2 = 4px), the focus is located at the point ((p, 0)). The primary focal chord is unique in that it is perpendicular to the axis of symmetry of the parabola and is the longest chord that can be drawn through the focus.
What is the equation of the parabola with focuse (07) and the directrix y1?
To find the equation of the parabola with focus at (0, 7) and directrix ( y = 1 ), we first determine the vertex, which is the midpoint between the focus and the directrix. The vertex is at ( (0, 4) ). The distance from the vertex to the focus is 3, so the parabola opens upward. The equation of the parabola can be expressed as ( (x - h)^2 = 4p(y - k) ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Thus, the equation is ( x^2 = 12(y - 4) ).
How do you write an equation of a parabola with vertex at the origin and the given focus 60?
To write the equation of a parabola with its vertex at the origin (0, 0) and a focus at (0, 60), you first identify the orientation of the parabola. Since the focus is above the vertex, the parabola opens upwards. The standard form of the equation for a parabola that opens upwards is ( y = \frac{1}{4p}x^2 ), where ( p ) is the distance from the vertex to the focus. Here, ( p = 60 ), so the equation becomes ( y = \frac{1}{240}x^2 ).
