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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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12y ago

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