Question

graph the function. compare to the graph of $f(x) = x^2$ (see examples 1 and 2).

1. $t(x) = -3x^2 + 1$

$x$

error analysis

1. describe and correct the error in graphing and comparing $f(x) = x^2$ and $g(x) = x^2 - 10$.

both graphs open up and have the same axis of symmetry. however, the vertex of the graph of $g$, $(0, 10)$, is 10 units above the vertex of the graph of $f$, $(0, 0)$.

Explanation:

Problem 10: $t(x) = -3x^2 + 1$

Step1: Choose x-values

Select $x = -2, -1, 0, 1, 2$

Step2: Calculate $t(x)$ values

For $x=-2$: $t(-2) = -3(-2)^2 +1 = -3(4)+1 = -11$
For $x=-1$: $t(-1) = -3(-1)^2 +1 = -3(1)+1 = -2$
For $x=0$: $t(0) = -3(0)^2 +1 = 0+1 = 1$
For $x=1$: $t(1) = -3(1)^2 +1 = -3(1)+1 = -2$
For $x=2$: $t(2) = -3(2)^2 +1 = -3(4)+1 = -11$

Step3: Plot points and compare

Plot $(-2,-11), (-1,-2), (0,1), (1,-2), (2,-11)$. Compared to $f(x)=x^2$: reflected over x-axis, vertically stretched by 3, shifted up 1 unit.

The error is in the direction of the vertex shift. For $g(x)=x^2 - 10$, the constant term $-10$ shifts the vertex of $f(x)=x^2$ downward 10 units, not upward. The vertex of $g(x)$ is $(0,-10)$, which is 10 units below $f(x)$'s vertex $(0,0)$.

Answer:

$x$	$-2$	$-1$	$0$	$1$	$2$

The graph of $t(x)$ is a downward-opening parabola (reflected over x-axis from $f(x)=x^2$), narrower (vertical stretch by factor 3), with vertex at $(0,1)$ (shifted up 1 unit from $f(x)$'s vertex $(0,0)$).

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Problem 18: Error Analysis for $f(x)=x^2$ and $g(x)=x^2-10$