Plotting the points (0, 2), (1, 4) and (–1, 0) and drawing a line passing through these points, we obtain the graph of of the line 2x – y 2 = 0 The shaded region represents the area bounded byVerified by Toppr y=2x 25x On factoring we get, x(2x5)=0 The roots are x=0,− 25 Since the constant term yintercept is 0, this graph passes through the origin//socraticorg/questions/whatisthevertexformofy2x22x6 y = 2(x 21)2 211 Explanation the equation of a parabola in vertex form is (∣∣∣∣∣(22)(y = a(x−h)2 k)(22)∣∣∣∣∣)
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F(x)=2x^2+8x+6 graph
F(x)=2x^2+8x+6 graph-First we find a tangent line of function y^2=16−2x at T(6,2) yt−y0=f′(x0)(x−x0)where x0=6,y0=2 y^2=16−2x⇒y=√(162x) or y=√(162x) Derivative of y=√(16−2x) at x0=6 is f′(x0)=−1/√(16 Since a straight line can be defined with as few as two points, you can do the following Plot the point (0,6) first, since the point where the line intercepts the y axis has x = 0
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Y = 2x2 − 6 y = 2 x 2 6 Find the properties of the given parabola Tap for more steps Direction Opens Up Vertex (0,−6) ( 0, 6) Focus (0,−47 8) ( 0, 47 8) Axis of Symmetry x = 0 x = 0Draw the graph of the equation, 2x y = 6 Find the coordinates of the point where the graph cuts the x axisAt what point the graph of 2xy=6, cuts xaxis?
Explanation one way is to find the intercepts, that is where the graph crosses the x and y axes ∙ let x = 0, in the equation for yintercept ∙ let y = 0, in the equation for xinterceptWe plot the points (0, 3) and (1, 8) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 5x 3 Let 2x 3y = 6 cut the yaxis at P therefore at PGraph y^2=2x y2 = 2x y 2 = 2 x Rewrite the equation as 2x = y2 2 x = y 2 2x = y2 2 x = y 2 Divide each term in 2x = y2 2 x = y 2 by 2 2 and simplify Tap for more steps x = y2 2 x = y 2 2 Find the properties of the given parabola Tap for more steps
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