How can I determine a equation from a graph? How can I determine how to graph y = x^2 + x - 2? Or y = x^2 - 4x + 4? Am I supposed to use the slope method (rise and run)?
I bet you have the zeroes from the graph. Look to see when the graph crosses the x axis, say at a and b.
then (x-a)(x=b)=0 mulitpy it out and you have it.
graphing y=x^2 -x-2=(x-2)(x+1)
it is zero (crosses x axis) at x=2, and x=-1. The max will be half way between those, or at x=.5
What is the max? Put in x=.5, calculate y. Graph those points
I tried the same method that you did to graph the equation y =x^2 -4x + 4 and my x-intercepts are both -2 (the y-intercept is y= 4). However, according to wolfram, the x-intercept should be 2. Who is correct: me or the website?
You did not factor correctly.
x^2-4x+4 = (x-2)(x-2)
If that's what you got, then you clearly don't understand what it means.
x-2 = 0 when x=2, not -2.
To determine an equation from a graph, you need to analyze the key features of the graph, such as the shape, intercepts, and any other points that might be evident. Let's consider the two given equations.
For the equation y = x^2 + x - 2:
1. Intercepts: To find the y-intercept, set x = 0 in the equation, giving us y = 0^2 + 0 - 2 = -2. Hence, the y-intercept is (0, -2). To find the x-intercepts, set y = 0, and solve the quadratic equation x^2 + x - 2 = 0. Factorizing or using the quadratic formula, we find the x-intercepts to be (-2, 0) and (1, 0).
2. Shape: From the equation y = x^2 + x - 2, we can infer that it represents a parabola, as it is a quadratic equation in the form y = ax^2 + bx + c.
3. Additional points: You could choose additional values of x to calculate the corresponding y values and plot those points on the graph to get a clearer picture.
For the equation y = x^2 - 4x + 4:
1. Intercepts: Similarly, set x = 0 to find the y-intercept. In this case, y = 0^2 - 4(0) + 4 = 4. Thus, the y-intercept is (0, 4). To find the x-intercepts, set y = 0, and solve the quadratic equation x^2 - 4x + 4 = 0. By factoring or using the quadratic formula, we find the x-intercept to be (2, 0).
2. Shape: This equation also represents a parabola, as it is a quadratic equation (y = ax^2 + bx + c).
3. Additional points: Once again, you can choose some x values, substitute them in the equation to calculate the corresponding y values, and plot those points.
Regarding the slope method (rise and run), it is not applicable in this case since we are dealing with quadratic equations and their graphs, which are not linear. The slope method is primarily used to determine slopes and identify the straight line equations (y = mx + b) from linear graphs.