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Mar 2, 2017 · The function has equation y = (1/2)^x. It may look like you only have 1 point, but in fact you have 2. Every exponential function has its y-intercept at (0, 1), because n^0 = 1 for every real value of n. We conclude the equation can be written in the form y = ab^x We can write a system of equations in two variables to solve for parameters a and b. {(8 = ab^-3), (1 = ab^0):} …

Apr 14, 2015 · Have a look at this, I hope it is what you need. If you have two ponts og given coordinates x_1, y_1, x_2,y_2:

Sep 27, 2014 · I assume that the grwoth factor same as is the growth constant. Let us look at an example. P (t) = P 0ekt, where P 0 is an initial value (= P (0)). To determine k, there should be another clue given like its doubling-life. Let us say the doubling-life is 3 years. P (3) = P 0ek(3) = 2P 0 by dividing by P 0, ⇒ e3k = 2 by taking the natural log, ⇒ lne3k = ln2 by simplifying a bit …

Feb 19, 2015 · I don't think you can make a model with the e or the two k and the t. However, you can make a y=ab^x model using the Exponential Regression key. To use this, go to STAT->CALC->and scroll down until you see "ExpReg". Then, if you have data entered into your STAT key, then you'll get a model for that. Hope that helps :)

How do you write an exponential function whose graph passes through the points (0,3) and (-1,6)?

Jan 21, 2015 · First, you input all your data points into your scatter plot (STAT->EDIT), and then turn on your plot (2ND->Y=->ENTER->ENTER). Once you have that done, go to your graph (Rightmost Key on the topmost row) and look at the points.

Jan 24, 2016 · By choosing suitable values for x and calculating the corresponding value of y . Then plot the points on squared paper. Finally draw a 'smooth' curve through them. x=0 : y = 20 = 1 ⇒ (0,1) x = 1 : y = 21 = 2 ⇒ (1,2) x = 2 : y = 22 = 4 ⇒ (2,4) x = 3 : y = 23 = 8 ⇒ (3,8) You can use negative values of x . x = − 1:y = 2−1 = 1 2 ⇒ (−1, 1 2) x = − 2:y = 2−2 = 1 4 ⇒ (−2, 1 4 ...

Apr 7, 2017 · All exponential functions can be written in the form y = abx. We just need to solve for a and b. Since the function goes through the points (1,6) and (2,12), x = 1 ⇒ y = 6 and x = 2 ⇒ y = 12 (1). Since x = 1 ⇒ y = 6, we just need to input these values into the …

Jul 14, 2015 · The standard formula for an exponential function is: f (x) = abx, with a ≠ 0 and b> 0 Given: f (1) = 6 and f (3) = 24 4 = 24 6 = f (3) f (1) = ab3 ab1 = b2 So b = ± √4 = ± 2. We want b> 0, so pick b = 2. Then a = ab1 b = f (1) b = 6 2 = 3 So a = 3 and b = 2, giving us f (x) = 3 ⋅ 2x General case: Given f (x1) = y1 and f (x2) = y2 bx2−x1 = abx2 abx1 = f (x2) f (x1) So b = (f (x2) f ...

May 13, 2016 · How do you find an exponential function of the form y = abx whose graph passes through the points (2,1) and (3,2)?