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3 years ago

What is the slop intercept form of a line that passes threw (4,-4) and (8, -10)

Mathematics

1 answer:

BigorU [14]3 years ago

4 0

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points through which the line passes, so we can find the slope:

$(x_ {1}, y_ {1}) :( 4, -4)\\(x_ {2}, y_ {2}) :( 8, -10)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-10 - (- 4)} {8-4} = \frac {-10+ 4} {4} = \frac {-6} {4} = - \frac {3} {2}$

Thus, the equation is of the form:

$y = - \frac {3} {2} x + b$

We substitute one of the points and find "b":

$-4 = - \frac {3} {2} (4) + b\\-4 = - \frac {12} {2} + b\\-4 + 6 = b\\b = 2$

Finally, the equation is of the form:

$y = - \frac {3} {2} +2$

ANswer:

$y = - \frac {3} {2} +2$

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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

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We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .