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x^2+3x is a quadratic function that may be factored into x*(x+3), which means that the graph has zeroes at x = 0 and x = -3. Because x^2, the dominant term, is positive, it means that the graph will be sloping up at the edges. It will look like a parabola that crosses the x-axis at 0 and -3 and opens up.

Thus, the equation of the axis of symmetry is x = -3/2 or -1.5, (which passes in the midway of the x-intercepts)), and the x-intercept of the vertex is -3/2. So replace x with -3/2 into the parabola's equation, and find the y-coordinate of the vertex, -9/4 or -2.25.

So that you have the vertex and the x-intercepts points, but still they are not enough to draw the parabola

Let x = 1, then y = 4. So you have two more points: (1, 4) and (-4, 4).

Plot these points and draw the parabola that passes through them.

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Wiki User

11y ago
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evelineclark

Lvl 2
3y ago

It's a parabolic curve y=X^2+3×=×*(×+3)

if x=0 => y=0

if y=0 => x=-3

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Anonymous

Lvl 1
4y ago

X^2+3×=×+(×+3)

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Q: How do you sketch the graph x squared plus 3x?
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