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Topic: **How to write and graph equations of lines****Question:**
2x + 5y = 0
-----------------------------------
Let x = 0
2(0) + 5y = 0
5y = 0
y=0
y intercept = (0,0)
----------------------------
Let y = 0
2x + 5(0) = 0
2x = 0
x = 0
x intercept = (0,0)
--------------------------------------...
In order to graph a line, I need two different coordinates of the line.
So what do I do?

June 15, 2019 / By Lauryn

The distance d between two points (x1, y1) and (x2, y2) is d = √[(x2 – x1)^2 + (y2 – y1)^2] Let (x,y) be a point on the graph of the function y = x^2. Use the distance formula to write an expression for the distance d between (x,y) and (3,0). [To do this I will replace x2 and y2 with x and y, and I’ll replace x1 and y1 with 3 and 0.] d = √[(x2 – x1)^2 + (y2 – y1)^2] d = √[(x – 3)^2 + (y – 0)^2] d = √[(x – 3)^2 + (y )^2] -----> simplify (y – 0) to y Now replace y with x^2 (since the function says y = x^2) and simplify. d = √[(x – 3)^2 + (x^2 )^2] d = √[(x – 3)^2 + (x )^4] d = √[(x ^2 - 6x + 9+ x^4] d = √[(x ^4 + x^2 - 6x + 9]

This is a tricky question. The distance formula is given by the following equation: d = sqrt[ (x2 - x1)^2 + (y2 - y1)^2 ] We want to know the distance from any point on the curve y = x^2 to the point (3,0). You need everything in the distance formula in terms of x, so in essence x2 remains as "x" in the domain of y = x^2 and x1 refers to the x-coordinate of your point, namely x1 = 3. y2 remains as "y=x^2" in the range and therefore y1 = 0 as in the point given. With all of that in mind you simply put in the information and begin simplifying: d = sqrt[ (x - 3)^2 + (x^2 - 0)^2] = sqrt[ x^2 -6x + 9 + x^4] so d(x) = sqrt[x^4 + x^2 - 6x + 9]<-----------------This is your function.

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