Given:
The image of a lens crosses the x-axis at –2 and 3.
The point (–1, 2) is also on the parabola.
To find:
The equation that can be used to model the image of the lens.
Solution:
If the graph of polynomial intersect the x-axis at c, then (x-c) is a factor of the polynomial.
It is given that the image of a lens crosses the x-axis at –2 and 3. It means (x+2) and (x-3) are factors of the function.
So, the equation of the parabola is:
...(i)
Where, k is a constant.
It is given that the point (–1, 2) is also on the parabola. It means the equation of the parabola must be satisfy by the point (-1,2).
Putting
in (i), we get



Divide both sides by -4.


Putting
in (i), we get

Therefore, the required equation of the parabola is
.
Note: All options are incorrect.
bearing in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient

![\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-7%3D1%5Bx-%28-1%29%5D%5Cimplies%20y-7%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%20y%3Dx%2B8%5Cimplies%20%5Cboxed%7B-x%2By%3D8%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bstandard%20form%7D%7D%7Bx-y%3D-8%7D)
just to point something out, is none of the options, however -x + y = 8, is one, though improper.
Answer:
2/3 of 27=18 km
Step-by-step explanation:
Answer:
Step-by-step explanation:
-x - 1 = x + 3
-2x - 1 = 3
-2x = 4
x = -2
y = -2 + 3
y = 1
(-2, 1)