(X +2)^2 - 5 = 0
X = - 4, Y = - 1
--------------------------- for intercepts
Just read it.
y = x2 +4x -1 (the standard form)
Since the vertex is (-2, -5), x = -2 is the equation of the axis of symmetry, also b= 4a (-b/2a = -2).
Since the parabola passes through (-4, -1) it also will passes through (0, -1), it means that c (y-intercept) equals to -1.
So we have,
y = ax2 + 4ax - 1 (replace x with -2, and y with -5)
-5 = 4a - 8a - 1
-4 = -4a
1 = a so that b = 4
To find the equation of a parabola with vertex at ((-3, 0)) that passes through the point ((3, 18)), we can use the vertex form of a parabola, (y = a(x + 3)^2). To determine the value of (a), substitute the point ((3, 18)) into the equation: [ 18 = a(3 + 3)^2 \implies 18 = a(6)^2 \implies 18 = 36a \implies a = \frac{1}{2}. ] Thus, the equation of the parabola is (y = \frac{1}{2}(x + 3)^2).
It is a parabola, which passes through the origin and is symmetric about the y axis.
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.
UndefinedImproved Answer:-The straight line equation is: y = 2x
y=x
please help
It is a parabola, which passes through the origin and is symmetric about the y axis.
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.
UndefinedImproved Answer:-The straight line equation is: y = 2x
y=13x
y=x
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
It passes through Quadrants II and IV. It also passes through the origin ... the point where the 'x' and 'y' axes cross. At that point, it's in all four quadrants.
y -4 = 3(x-3)y = 3x -5
Yes, it is a linear equation with no slope. The graph is a straight line (parallel to the x-axis) that passes through the y-axis, at {0,-3}
2
The equation is x = -7.