To graph the quadratic function ( y = -0.5x^2 + 2x + 1 ), first identify the vertex, which can be found using the formula ( x = -\frac{b}{2a} ). For this equation, ( a = -0.5 ) and ( b = 2 ), so the vertex occurs at ( x = 2 ). Calculate the corresponding ( y ) value to find the vertex point, and then plot additional points by substituting values for ( x ) into the equation. Finally, draw a smooth curve to represent the parabola, which opens downwards due to the negative coefficient of ( x^2 ).
If you mean: 2x+4y = 4 then the graph joins the points: (2, 0) and (0, 1)
-2x plus 3y equals 1
-2x+3y=1 3y=1+2x y=(1+2x)/3 Then proceed to find points by plugging in given or arbitrary values of x.
2x + 6y = 6Subtract 2x from each side:6y = -2x + 6Divide each side by 6:y = -1/3 x + 1The graph is a straight line, with a slope of [ -1/3 ], and intersecting the y-axis at the point [ y = 1 ].
Flipping the graph of the function ( y = x^2 + 2x - 2 ) vertically involves multiplying the entire function by -1. This results in the new equation ( y = -(x^2 + 2x - 2) ), which can be simplified to ( y = -x^2 - 2x + 2 ). So, yes, the flipped graph can be represented as ( y = -(x^2 + 2x - 2) ).
If you mean: 2x+4y = 4 then the graph joins the points: (2, 0) and (0, 1)
3
-2x plus 3y equals 1
-2x=2y+5 +2x -2y -2y=2x+5 /-2 y=-1/1+2.5
-2x+3y=1 3y=1+2x y=(1+2x)/3 Then proceed to find points by plugging in given or arbitrary values of x.
Any line which has a gradient which is not 2 will not be parallel to the line y = 2x + 1.
2x + 6y = 6Subtract 2x from each side:6y = -2x + 6Divide each side by 6:y = -1/3 x + 1The graph is a straight line, with a slope of [ -1/3 ], and intersecting the y-axis at the point [ y = 1 ].
the equation should be y =-4/2x+1 that is a way to make it easier to graph
Flipping the graph of the function ( y = x^2 + 2x - 2 ) vertically involves multiplying the entire function by -1. This results in the new equation ( y = -(x^2 + 2x - 2) ), which can be simplified to ( y = -x^2 - 2x + 2 ). So, yes, the flipped graph can be represented as ( y = -(x^2 + 2x - 2) ).
1
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
As stated these are not lines, but just a collection of algebraic terms. If we change them to y=2x and y=2x-1, then on a graph of y versus x, these are parallel lines separated by vertical distance of 1.