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Solving Exponential Equations Using Log Laws>
Solving Exponential EquationsSolving Exponential Equations
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Question 1 of 2
1. Question
Find the graph for `y=(x -2)^3+3` by using `y=x^3`.
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Horizontal and vertical translations (shifts) of cubic functions are written in the form `y=(x-color(red)(h))^3 +color(blue)(c)` where the point `(color(red)(h),color(blue)(c))` is the inflection point of the function.`color(red)(-h)` is a shift to the right and `color(blue)(+c)` is a shift upwards.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(darr)` Shift Down`(+c) \ bb(uarr)` Shift UpTo obtain the graph of `y=(x -color(red)(2))^3+color(blue)(3)`, first sketch the function `y=x^3`.Sketch the function `y=x^3`. Remember the formula `y=(x-h)^3 +c` when applied to `y=x^3` (can be rewritten as `y=(x-color(red)(0))^3+color(blue)(0)`) has its inflection point at `(color(red)(0),color(blue)(0))`.
Using the formula `y=(x-h)^3 +c` for horizontal and vertical translations (shifts) and remembering that the point `(h,c)` is the inflection point of the function, the inflection point for `y=(x -color(red)(2))^3+color(blue)(3)` is `(color(red)(2),color(blue)(3))`.
Sketch the curve for `y=(x -color(red)(2))^3+color(blue)(3)` through its inflection point `(color(red)(2),color(blue)(3))` following the same shape as `y=x^3`.
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Question 2 of 2
2. Question
Find the graph for `y=x^2 – 6x + 11` by using `y=x^2`.
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Horizontal and vertical translations (shifts) of quadratic functions are written in the form `y=(x-color(red)(h))^2 +color(blue)(c)` where the point `(color(red)(h),color(blue)(c))` is the vertex of the function.`color(red)(-h)` is a shift to the right and `color(blue)(+c)` is a shift upwards.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(darr)` Shift Down`(+c) \ bb(uarr)` Shift UpTo obtain the graph of `y=x^2 – 6x + 11`, first sketch the function `y=x^2`.Sketch the function `y=x^2`. Remember the formula `y=(x-color(red)(h))^2 +color(blue)(c)` when applied to `y=x^2` (can be rewritten as `y=(x-color(red)(0))^2+color(blue)(0)`) has its vertex at `(color(red)(0),color(blue)(0))`.
Then rewrite `y=x^2 – 6x + 11` in the standard form by completing the square.`y=(x^2-6x+9)-9+11``y=(x -3)^2+2`.Remember the standard form is `y=(x-h)^2 +c`.Using the formula `y=(x-h)^2 +c` for horizontal and vertical translations (shifts) and remembering that the point `(h,c)` is the vertex of the function, the vertex point for `y=(x -color(red)(3))^2+color(blue)(2)` is `(color(red)(3),color(blue)(2))`.
Sketch the curve for `y=(x -color(red)(3))^2+color(blue)(2)` through its vertex point `(color(red)(3),color(blue)(2))` following the same shape as `y=x^2`.