How do you find the point slope form? (given 2 points)
2 answers:
Answer and Step-by-step explanation:
First, define point-slope form. Point-slope form is a format of linear equations, written as: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point.
Let's say we have two points (a, b) and (c, d). We need to find the slope. Remember that slope is change in y divided by change in x. "Change in y" basically means the difference in the y-coordinates. Here, it would be: d - b. "Change in x" means the difference in the x-coordinates. Here, it would simply be c - a. So, the slope would be: m = .
And, since we have two points, we can choose any of them to be our "(x1, y1)" point to use. Arbitrarily, I'm just going to choose (a, b), but you could do it with (c, d), as well. So, a = x1 and b = y1, and plug these in:
y - y1 = m(x - x1)
And, there's your equation is in point-slope form!
You might be interested in
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i
well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too
so, the roots are -2, 2i, -2i
now...![\bf \begin{cases} x=-2\implies x+2=0\implies &(x+2)=0\\ x=2i\implies x-2i=0\implies &(x-2i)=0\\ x=-2i\implies x+2i=0\implies &(x+2i)=0 \end{cases} \\\\\\ (x+2)\underline{(x-2i)(x+2i)}=0\\\\ -----------------------------\\\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ (x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0 \\\\\\ (x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0 \\\\\\ x^3+2x^2+4x+8=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%3D-2%5Cimplies%20x%2B2%3D0%5Cimplies%20%26%28x%2B2%29%3D0%5C%5C%0Ax%3D2i%5Cimplies%20x-2i%3D0%5Cimplies%20%26%28x-2i%29%3D0%5C%5C%0Ax%3D-2i%5Cimplies%20x%2B2i%3D0%5Cimplies%20%26%28x%2B2i%29%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B2%29%5Cunderline%7B%28x-2i%29%28x%2B2i%29%7D%3D0%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bdifference%20of%20squares%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%28a-b%29%28a%2Bb%29%20%3D%20a%5E2-b%5E2%5Cqquad%20%5Cqquad%20%0Aa%5E2-b%5E2%20%3D%20%28a-b%29%28a%2Bb%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28x%2B2%29%5Bx%5E2-%282i%29%5E2%5D%3D0%5Cimplies%20%28x%2B2%29%5Bx%5E2-%282%5E2i%5E2%29%5D%3D0%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B2%29%5Bx%5E2-%284%5Ccdot%20-1%29%5D%3D0%5Cimplies%20%28x%2B2%29%28x%5E2%2B4%29%3D0%0A%5C%5C%5C%5C%5C%5C%0Ax%5E3%2B2x%5E2%2B4x%2B8%3D0)
now, if we check f(-1), we end up with 5, not 15
hmmm
so, how to turn our 5 to 15? well, 3*5, thus
usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too
so, the roots are -2, 2i, -2i
now...
now, if we check f(-1), we end up with 5, not 15
hmmm
so, how to turn our 5 to 15? well, 3*5, thus
usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15