You need graph paper upon which you draw 2 lines at right angles. These are called the "axes" (singular is axis). The x axis should be near the centre of the graph paper and running horizontally; the y axis should be near the centre of the graph paper and running vertically.
Now you have to put a scale along each axis; the x axis should run left to right from negative numbers to positive numbers;the y axis should run from negative numbers at the bottom part and positive numbers above the "origin".
The origin is where the axes cross where the two scales read zero. That is the point (0,0); x numbers written first.
Then you have to compile a table of "values" of y for lots of separate values of x
Example sfor y = x^2 + 4x + 3 (where the sign ^ means "to the power of"
If x = 0, y = 3;
If x = 1, y = 1^2 + (4 x 1) + 3 = 1 + 4 + 3 = 8
If x = 2, y = 2^2 + (4 x 2) + 3 = 4 + 8 + 3 = 15
So we have found the x and y values for 3 points so far; they are (0,3); (1,8) and (2,15) and these can now be "plotted" onto the graph paper, remembering that the x value is the left hand one of each pair.
To plot the point (3, 24) you would go +3 on the x axis and "up" as far5 as 24 on the y axis.
As you keep on calculating more values of x for values of y you will be able to plot many more points; the more you plot the more you will be able to see the shape of the curve which represents y = x^2 + 4x + 3
or
y = x2 + 4x + 3
Let x = 0, then (0, 3) is the y-intercept point (plot it).
Let y = 0, then (x + 3)(x +1) = 0 so that (-3, 0) and (-1, 0) are the x-intercepts points (plot them).
Since -2 is midway -3 and -1, x = -2 is the equation of the axis of symmetry. So you can plot (-4, 3) as it is the symmetric point of (0,3).
Since the vertex lies on the axis of symmetry, -2 is the x-coordinate of the vertex, and its y-coordinate it will be -1, as you replace -2 with x into the equation. So go and plot (-2, -1), the vertex.
Now draw a smooth curve that passes through the points that you've just plotted. And you're done, that is your parabola.
It is the equation of a parabola.
oh my god, that was the exact homework i had few days ago, but you have to have a calculator to do this. just plug in the numbers in the but ton Y=, one column you put-16t squared plus 25t plus 6 then the next column you put 0. the graph should form a parabola.
You are describing a circle, with its center at the origin and a radius of 4 (the square root of 16)
104
2 squared plus 2 x 3 = 10, 7 squared plus 7 x 2 = 63, 6 squared plus 6 x 5 = 66,8 squared plus 8 x 4 = 96 so 9 squared plus 9 x 7 = 81 + 63 = 144.
It is the equation of a parabola.
(-3, -5)
The graph is a circle with a radius of 6, centered at the origin.
7
The minimum value of the parabola is at the point (-1/3, -4/3)
It is a parabola with its vertex at the origin and the arms going upwards.
oh my god, that was the exact homework i had few days ago, but you have to have a calculator to do this. just plug in the numbers in the but ton Y=, one column you put-16t squared plus 25t plus 6 then the next column you put 0. the graph should form a parabola.
The graph is a circle with a radius of 6, centered at the origin
The equation y = 4x^2 + 5 is a parabola
(6, 40) and (-1, 5)
y=xsquared-4x+2
A quadratic equation (t=s2+3). This kind of line will result in a parabola like curve.