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When you are given 2 points, like (-2,2) and (2,162) How can you tell which equation contains those points? Do I have to change and put values of y and x and solve? Thanks

3 answers

To determine which equation contains the points \((-2, 2)\) and \((2, 162)\), you can use the following methods:

  1. Slope-Intercept Method: If you are given a general linear equation in the form of \(y = mx + b\), you can calculate the slope (\(m\)) between the two points and then use one of the points to find the y-intercept (\(b\)).

    • Calculate the slope \(m\): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{162 - 2}{2 - (-2)} = \frac{160}{4} = 40 \]

    Now that you have the slope, use one of the points to find \(b\). Let's use point \((-2, 2)\): \[ y = mx + b \implies 2 = 40(-2) + b \implies 2 = -80 + b \implies b = 82 \]

    Thus, the equation is: \[ y = 40x + 82 \]

  2. Checking against potential equations: If you have multiple choice equations, you can substitute both \(x\) and \(y\) values from each point into the equation. If both points satisfy the equation, then that equation contains both points.

For example, if you have an equation like \(y = 40x + 82\), you can check:

  • For \((-2, 2)\): \[ y = 40(-2) + 82 = -80 + 82 = 2 \quad \text{(True)} \]
  • For \((2, 162)\): \[ y = 40(2) + 82 = 80 + 82 = 162 \quad \text{(True)} \]

Thus, both points lie on the line described by the equation \(y = 40x + 82\).

  1. Using the standard form or other forms: If you are given equations in different forms (standard, point-slope, etc.), you can also substitute points directly to determine which equation fits.

In summary, you can indeed put the values of \(x\) and \(y\) into potential equations to solve if you have specific options, or you can derive the equation as shown above using the slope.

If it is an exponential equation, do I use substitution too to find out? Thanks

Yes, if you suspect that the relationship between the two points is exponential rather than linear, you can still use substitution to determine if a given exponential equation contains the points. Here’s how to approach it:

General Form of an Exponential Equation

Exponential equations typically take the form: \[ y = ab^x \] where \(a\) is a constant, \(b\) is the base of the exponential (where \(b > 0\) and \(b \neq 1\)), and \(x\) is the exponent.

Finding the Equation

To find the specific exponential equation that fits the given points:

  1. Substitute the Points into the General Form: You have two points: \((-2, 2)\) and \((2, 162)\).

    Using Point \((-2, 2)\): \[ 2 = ab^{-2} \quad \text{(1)} \]

    Using Point \((2, 162)\): \[ 162 = ab^{2} \quad \text{(2)} \]

  2. Solve the System of Equations: You now have a system of two equations. From equation (1), you can express \(a\) in terms of \(b\): \[ a = 2b^{2} \quad \text{(from equation 1)} \]

    Substitute \(a\) into equation (2): \[ 162 = (2b^{2})b^{2} = 2b^{4} \]

    Now, divide both sides by 2: \[ b^{4} = 81 \]

    Taking the fourth root: \[ b = 3 \quad \text{(since we need a positive base)} \]

  3. Find \(a\): Substitute \(b\) back into the first equation to find \(a\): \[ 2 = ab^{-2} \implies 2 = a \cdot 3^{-2} \implies 2 = \frac{a}{9} \implies a = 18 \]

  4. Write the Exponential Equation: With \(a\) and \(b\), you can write the exponential equation: \[ y = 18 \cdot 3^x \]

Checking the Given Points

Now you can check to see if the original points satisfy this equation:

  • For \((-2, 2)\): \[ y = 18 \cdot 3^{-2} = 18 \cdot \frac{1}{9} = 2 \quad \text{(True)} \]

  • For \((2, 162)\): \[ y = 18 \cdot 3^{2} = 18 \cdot 9 = 162 \quad \text{(True)} \]

Conclusion

If you have specific candidate exponential equations, you would substitute the points into those equations to see if both points satisfy the same equation. If the calculations yield true statements for both points, then the equation contains those points.