Question

what's the full equation for this graph?

Answer 1

1.
Left side of a graph is a line that passes through points $A=(-1,-1)$ and $B=(1,4)$ so we can write its equation as:

$(y-y_A)=\dfrac{y_B-y_A}{x_B-x_A}(x-x_A)\\\\\\ \big(y-(-1)\big)=\dfrac{4-(-1)}{1-(-1)}\big(x-(-1)\big)\\\\\\ y+1=\dfrac{4+1}{1+1}(x+1)\\\\\\y+1=\dfrac{5}{2}(x+1)\\\\\\y+1=\dfrac{5}{2}x+\dfrac{5}{2}\\\\\\y=\dfrac{5}{2}x+\dfrac{5}{2}-1\\\\\\\boxed{y=\frac{5}{2}x+\frac{3}{2}}$

2.
Right side of a graph is a parabola with vertex $V=(3,-3)$ that passes through point $P=(1,1)$. Its equation:

$y=a(x-x_V)^2+y_V\\\\y=a(x-3)^2+(-3)\\\\y=a(x-3)^2-3$

Substitute $x=1$ and $y=1$ to calculate parameter $a$

$1=a(1-3)^2-3\\\\1=a(-2)^2-3\\\\1=4a-3\\\\4a=4\quad|:4\\\\a=1$

and we have the equation of parabola:

$y=a(x-3)^2-3\\\\y=1\cdot(x-3)^2-3\\\\y=(x-3)^2-3\\\\y=x^2-2\cdot x\cdot3+3^2-3\\\\y=x^2-6x+9-3\\\\\boxed{y=x^2-6x+6}$

From 1. and 2. full equation for graph is:

$f(x)=\begin{cases}\dfrac{5}{2}x+\dfrac{3}{2}\qquad\text{for}\,\,\,x\ \textless \ 1\\\\x^2-6x+6\qquad\text{for}\,\,\,x\geq1\end{cases}$