How do you identify the important parts of $y = x^{2} - 1$ $y=x^2-1$ to graph it?

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How do you identify the important parts of $y = x^{2} - 1$ $y=x^2-1$ to graph it?

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Answer 1 · 2015-10-19T03:51:30

In summary, find the vertex and x-intercepts, and plug and chug for additional points. Finally, connect 'em all together with a neat curve.

Explanation:

The "important parts" in terms of graphing would be the x-intercepts and the vertex. From there you can just plug and chug to identify other points on the graph.

To find the x-intercepts, you set $y = 0$ $y = 0$ and solve for your $x$ $x$ s:
$0 = x^{2} - 1$ $0 = x^2-1$
$1 = x^{2}$ $1 = x^2$ (Adding 1 to both sides)
$x = \pm 1$ $x = +-1$ (Taking square roots)

Thus, the x-intercepts occur at $x = 1$ $x = 1$ and $x = - 1$ $x = -1$ .

The vertex is the "beginning" point of a quadratic like this one. In other words, it's where the two curves meet on a parabola (not a very good definition, but it goes). To find the x-coordinate of the vertex, we use the formula $x = - \frac{b}{2 a}$ $x = -b/(2a)$ , where $a$ $a$ and $b$ $b$ are the numbers in $a x^{2} + b x + c$ $ax^2+bx+c$ (which is the form of quadratic presented in this problem). In our equation $x^{2} - 1$ $x^2-1$ , $a = 1$ $a = 1$ and $b = 0$ $b = 0$ (since there is no middle term, we have $b = 0$ $b = 0$ ). Thus, $x = - \frac{0}{2 (1)} = 0$ $x = -0/(2(1)) = 0$ . This is the x-coordinate of the vertex; to find the y, we simple plug in: $y = x^{2} - 1 = {(0)}^{2} - 1 = - 1$ $y = x^2-1 = (0)^2-1 = -1$ . The coordinates of the vertex are $(0, - 1)$ $(0,-1)$ .

We can see everything above on the graph below. The vertex, $(0, - 1)$ $(0,-1)$ , is the bottom of the graph; you'll want to start there and work your way up. Next are the intercepts at -1 and 1 on the x-axis; plot those points, and connect them (vertex, and both intercepts) together with a nice curve. Finally, choose x-values like -3, -2, 2, and 3 and find which y-values they produce:
$y = {(- 3)}^{2} - 1 = 9 - 1 = 8$ $y = (-3)^2-1 = 9-1 = 8$
$y = {(- 2)}^{2} - 1 = 4 - 1 = 3$ $y = (-2)^2-1 = 4-1 = 3$
$y = {(2)}^{2} - 1 = 4 - 1 = 3$ $y = (2)^2-1 = 4-1 = 3$
$y = {(3)}^{2} - 1 = 9 - 1 = 8$ $y = (3)^2-1 = 9-1 = 8$
And just like that, we've identified 4 more points to plot:
$(- 3, 8)$ $(-3,8)$
$(- 2, 3)$ $(-2,3)$
$(2, 3)$ $(2,3)$
$(3, 8)$ $(3,8)$

Finally finally, to complete the rough sketch of the graph, connect all the points together - the "special" ones (vertex and intercept(s)) and the ones we found by plugging and chugging.

graph{x^2-1 [-10, 10, -5, 5]}

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How do you identify the important parts of $y = x^{2} - 1$ $y=x^2-1$ to graph it?

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