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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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0It's a parabolic curve y=X^2+3×=×*(×+3)

if x=0 => y=0

if y=0 => x=-3

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5x2 + 3x + 8x2 = 13x2 + 3x = x(13x + 3)

The given expression can be simplified to: 3X squared +27

3x squared

(3x + 3y)(x + y) = 3(x + y)2

(3x - y+ 2)(3x + y + 2)

It is: 3x*(x+7)

The given expression can be simplified to: 8x squared +9

(3x + 1)(x + 5)

(x + 5)(3x + 1)

3(x2 + 3x + 1)

x^2 + 3x + 2/(-3x) - x^2 - 2=3x-2+2/(-3x)=3x-2-2/(3x)

12

If that's -18y2, the answer is (3x - 2y)(x + 9y)

(6x + t)(3x + t)

6xyz(3x + 2y + z)

Rearrange the equation to read : y = 6 - 3x and graph normally.

3x(2x + 1)

3x2 + 4x factors to x(3x + 4)

15x squared plus eleven

9x2...I guess that means 9x squared ? The problem: 9x(squared)+12x+4 Here's the answer: (3x+2)(3x+2) or even more simplified: (3x+2) squared

3x2 + x = x*(3x + 1)

-27

To graph y2 = -3x, first solve for y. Doing this results in two solutions: y = √(-3x) and y = -√(-3x) Put the first solution into y1 and the second solution into y2. The two solutions together should form a sideways parabola.

3x2 + 10x + 3 = (x + 3)(3x + 1).

It is a quadratic expression and can be factored into: (3x+1)(x+2)