Teresa graphs the following 3
equations: y=2x
, y=x2+2
, and y=2x2
.
She says that the graph of y=2x
will eventually surpass both of the other graphs.
Is Teresa correct? Why or why not?
Caption: Graph with X-axis labeled at negative 5 and 5 and Y-axis labeled at negative 3 and 5. Lat: 3 equations are graphed in different colors. Equation 1 is y = 2 superscript x and is a curve opening up and to the left, starting near the X-axis at left and continuing up and ending near the top right of the graph. The curve gets steeper toward the right. Equation 2 is y = 2 x squared and is a parabola opening upward with vertex (0, 0) that passes through (negative 1, 2) and (1, 2). Equation 3 is y = x squared + 2, a wider parabola opening upward with vertex (0, 2) that passes through (negative 2, 6) and (2, 6).
CLEAR CHECK
Teresa is correct.
The graph of y=2x
grows at an increasingly increasing rate, but the graphs of y=x2+2
and y=2x2
both grow at a constantly increasing rate.
Therefore, the graph of y=2x
will eventually surpass both of the other graphs.
Teresa is not correct.
The graph of y=2x
grows at an increasing rate and will eventually surpass the graph of y=x2+2
.
However, it will never surpass the graph of y=2x2
because the y
-value is always twice the value of x2
.
Teresa is not correct.
The graph of y=2x2
already intersected and surpassed the graph of y=2x
at x=1
.
Once a graph has surpassed another graph, the other graph will never be higher.
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1 answer
The graph of y=2x will eventually surpass the graph of y=x^2+2, as it grows at an increasing rate. However, it will never surpass the graph of y=2x^2, as the y-value in this equation is always twice the value of x^2.