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Now, let's get back to parabolas. There is a basic procedure we can always use to get a pretty good sketch of a parabola. Following it is.
1. Determine the vertex. We'll discuss how to determine this shortly. It's quite simple, but there are several methods for finding it and so will be discussed separately.
2. Find the y-intercept, (0, f (0)) .
3. Solve f ( x ) = 0 to determine the x coordinates of the x-intercepts if they exist.
4. Ensure that you've got at least one point to either side of the vertex. It is to ensure we get a somewhat accurate sketch. If the parabola contains two x-intercepts then already we'll have these points. If it contains 0 or 1 x-intercept we can either just plug in another x value or employ the y-intercept and the axis of symmetry to obtain the second point.
5. Sketch the graph. At this point we've gotten sufficient points to get a quite decent idea of what the parabola will look like.
Simpler method Let's begin by looking at the simpler method. This method will employ the following fact about exponential functions. If b x = b y then x
use substitution method to solve this equation x+y=20 y= -5x
You need to prepare a 30 mL solution of a 1:6 syrup solution. You have on hand a 50% syrup solution and a 1:200 soda solution. How many mL of each solution will you need?
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3^x + 3^(x^1/2) - 90 =0
Which shows the quadratic function f(x) = 2(x - 4)2 - 7 in standard form?
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Logarithm form In this definition y = log b x is called the logarithm form Exponential form In this definition b y = x is called the exponential form.
3x+y=-2 6x+2y=10
how do you graph linear equations?
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