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Assuming the first function is y = 4x2 + 1, and the second y = 4x2 then the first graph is 1 unit higher than the second.
Their noses are both at the origin, and they both open upward, but y=4x2 is a much skinnier parabola.
4
if the linear equation is x+y=1 means we are having the graph points (1,0) (2,-1)....using this graph we can draw the graph
The function y = x is the graph that passes from the points (-1, -1), (0, 0), and (1, 1) The function y = 4x is the graph that passes form the points (-1, -4), (0, 0), and (1, 4) Sketch these graphs in a same x and y coordinate system, and you can see both of them
3
Their noses are both at the origin, and they both open upward, but y=4x2 is a much skinnier parabola.
y = -4x2 + 1 Range = {y:y=<1, y is an element of the reals}
No.
The equation y = 4x^2 + 5 is a parabola
It is a parabola with its vertex at the origin and the arms going upwards.
yes
4x2-y2 = (2x-y)(2x+y)
In real numbers there is no square root of -4x2 However, when using complex numbers where √-1 = i, then: √-4x2 = 2xi If you were asking, for example, with regard to the difference of two squares, eg y2 - 4x2, then you are actually finding the square root of 4x2 which is 2x, giving: y2 - 4x2 = (y + 2x)(y - 2x)
y = 4x2 + 6 ory = 4x2 + 0x + 6, where a = 4, b = 0 and c = 6x-coordinate of the vertex = -b/2a = -0/2(4) = 0y-coordinate of the vertex = 4(02) + 6 = 6vertex is (0, 6)(Or you can say: Since the vertex of y = x is (0, 0), and this graph is shifted 6 units upward in the y-axis, then the vertex of y = x + 6 would be (0, 0 + 6) or (0, 6)Substitute 0 for y in order to find the roots.y = 4x2 + 60 = 4x2 + 60 - 6 = 4x2 + 6 - 6-6 = 4x2-6/4 = 4x2/4- 6/4 = x2± √(-6/4) = x± [√(6/4)]i = x x = [(1/2)√6]i or x = - [(1/2)√6]i
First you need to solve for y. So write 4x2+y=16 so y=16-4x2 Now write f(x)=16-4x2
y=4x2+3x+8
The slope or gradient is the same and would run parallel but the intercept on the y axis would be at -1