There are two (main) ways to graph this parabola.
The first is to simply substitute in values of x, find the corresponding y values and then plot those points and connect the points with a curve.
The second way is to 'complete the square' so that we can find where the maximum or minimum occurs and the y-intercept and graph from those two points. This is the process described below.
y = -x2 + 8x + 5
y = -(x2 - 8x) + 5
y = -(x2 - 8x + 16 - 16) + 5
y = -(x2 - 8x + 16) + 16 + 5
y = -(x-4)2 + 21
From this we can see that a maximum occurs at (4,21). To find the y-intercept, substitute in x=0, giving y=5.
From these two points, we get a pretty good idea of what the graph looks like. If you want more accuracy, substitute in more values of x.
y=x2+4x+1
The graph is a parabola facing (opening) upwards with the vertex at the origin.
The vertex has a minimum value of (-4, -11)
Y=X^2 is a function for it forms a parabola on a graph.
It looks like a parabola which looks like a U shape.
y=x2+4x+1
9
The graph is a parabola facing (opening) upwards with the vertex at the origin.
The vertex has a minimum value of (-4, -11)
Y=X^2 is a function for it forms a parabola on a graph.
It looks like a parabola which looks like a U shape.
Y = X2 ===== The graph of this parabola is crossed only at a point and once by a vertical line, so it is a function. Passes the vertical line test.
Go to the y= button at the top left corner, right below the screen. Then, type in "x^2" or type X, then use the X2 button. Then, press Graph on the top right of the calculator. The y=x2 function is the most basic parabola.
Instead of the answer being a curve, it is a region. For example, if y > x2 + 4, the answer is not the parabola y = x2 + 4. Instead it is the region above the parabola (as if the bowl were filled with something.)
Their noses are both at the origin, and they both open upward, but y=4x2 is a much skinnier parabola.
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
The cubic function is the name of graph that is steeper on the way up than on the way down. An absolute value function is a grpah that is shaped a bit like a y=x2 parabola.