Graph the parabola.

y=-2x^2+12x-16

Bot GPT 3.5 Bot GPT 3.5 answered
5 months ago

To graph the parabola y = -2x^2 + 12x - 16, we can start by finding the vertex.

The x-coordinate of the vertex can be found using the formula x = -b/2a, where a = -2 and b = 12.
Substituting these values, we get x = -12/(2*-2) = -12/(-4) = 3.

To find the y-coordinate of the vertex, we can substitute x = 3 into the equation:
y = -2(3)^2 + 12(3) - 16
= -2(9) + 36 - 16
= -18 + 36 - 16
= 2.

So the vertex of the parabola is (3, 2).

Next, we can find the y-intercept by substituting x = 0 into the equation:
y = -2(0)^2 + 12(0) - 16
= -2(0) + 0 - 16
= 0 - 16
= -16.

Therefore, the y-intercept is (0, -16).

To find another point on the graph, we can choose a value for x and calculate the corresponding y-value. For example, if x = 1:
y = -2(1)^2 + 12(1) - 16
= -2(1) + 12 - 16
= -2 + 12 - 16
= -6.

So the point (1, -6) is also on the graph.

Using these points, we can sketch the graph:

Firstly, plot the vertex at (3, 2).
Then, plot the y-intercept at (0, -16).
Finally, plot the point (1, -6).

The graph of the parabola y = -2x^2 + 12x - 16 is a downward opening parabola that passes through these three points.

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