Question

How can I find where parobala intersect

Answer 1

`\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\\\ % left side templates \begin{array}{llll} f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ y=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}} \end{array}\\\\ --------------------\\\\`


`\bf % template detailing \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the y-axis}`


`\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by }{{ D}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}}`

so.. with that template in mind, let's do the 4units shift, by setting C = +4,
and leaving B = 1, since C/B = +4/1 = +4


`\bf y=(x-2\boxed{+4})^2+3\implies y=(x+2)^2+3\impliedby \textit{shifted version}`

so.. now... where do they meet? well, is just a system of equations of two variables, and they meet when both equations equal each other

since both equations are solved by "y", let's use substitution


`\bf \begin{cases} y=(x-2)^2+3\\ y=(x+2)^2+3\\ ----------\\ (x-2)^2+3=(x+2)^2+3 \end{cases} \\\\\\ (x-2)^2=(x+2)^2\implies x^2-4x+4=x^2+4x+4 \\\\\\ -4x=4x\implies 0=8x\implies \boxed{0=x} \\\\\\ \textit{let's plug that in the first equation, to get`