Now, make one final coaster that has the works! Include each of the following: A "swoop" down through the x-axis at x = 300. (Th
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Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is
<u>Solution:</u>
Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).
Generic slope intercept form of a line is given by y = mx + c
where m = slope of the line.
Let's first find slope intercept form of 3x + y = -8
3x + y = -8
=> y = -3x - 8
On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3
And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.
Let required slope = x
So we need to find the equation of a line whose slope is and passing through (-3,1)
Equation of line passing through and having lope of m is given by
Substituting the values we get,
Hence the required equation of line is found using slope intercept form
Answer:
2/5
Step-by-step explanation:
Hi! Whenever you find a limit, you first directly substitute x = 5 in.
Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.
For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.
From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.
From x²-25, you can factor as (x+5)(x-5) via differences of two squares.
After factoring the expressions, we get a new Limit.
We can cancel x-5.
Then directly substitute x = 5 in.
Therefore, the limit value is 2/5.
L’Hopital Method
I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.
The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.
So from the given function:
Differentiate numerator and denominator, apply power rules.
<u>Differential</u> (Power Rules)
<u>Differentiation</u> (Property of Addition/Subtraction)
Hence from the expressions,
<u>Differential</u> (Constant)
Therefore,
Now we can substitute x = 5 in.
Thus, the limit value is 2/5 same as the first method.
Notes:
- If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.