## University Calculus: Early Transcendentals (3rd Edition)

- The green graph: $y =-(x-2)^2$ - The blue graph: $y=-(x-1)^2+4$ - The yellow graph: $y=-(x+4)^2-1$ - The red graph: $y=-(x+2)^2+3$
Here we would use the principles of graph shifting and examine each given graph to know how the graph has shifted from its original position and thus figure out its equation. The original equation here is $y = -x^2$. You can take a look back at the blue graph at exercise 21 to know which position the graph of $y=-x^2$ possesses. It collides the x-axis at $x=0$ and y-axis at $y=0$. The equation after shifiting would have the form $$y=-(x+a)^2+b$$ where $a$ and $b$ correspond to which direction and how many units the original graph has shifted. 1) The blue graph: The highest point (vertex) of the parabola is $(1,4)$, where $x=1$ and $y=4$. So we know: - This graph has shifted to the right 1 unit, meaning $a=-1$ - It also has shifted up 4 units, so $b=4$ Therefore, the equation of the blue graph is $$y=-(x-1)^2+4$$ 2) The red graph: The highest point (vertex) of the parabola is $(-2,3)$, where $x=-2$ and $y=3$. So we know: - This graph has shifted to the left 2 units, meaning $a=2$ - It also has shifted up 3 units, so $b=3$ Therefore, the equation of the red graph is $$y=-(x+2)^2+3$$ 3) The yellow graph: The highest point (vertex) of the parabola is $(-4,-1)$, where $x=-4$ and $y=-1$. So we know: - This graph has shifted to the left 4 units, meaning $a=4$ - It also has shifted down 1 unit, so $b=-1$ Therefore, the equation of the yellow graph is $$y=-(x+4)^2-1$$ 4) The green graph: The highest point (vertex) of the parabola is $(2,0)$, where $x=2$ and $y=0$. So we know: - This graph has shifted to the right 2 units, meaning $a=-2$ - It has not shifted vertically, so $b=0$ Therefore, the equation of the green graph is $$y=-(x-2)^2$$