To find where the parabola intersects the x -axis, we set y = 0. This gives us -x² + 3x = 0. Factoring out -x, we get -x(x - 3) = 0. This implies that x = 0 or x = 3. So, the parabola crosses the x -axis at (0, 0) and (3, 0).

To find where the parabola intersects the x -axis, we set y = 0. This gives us -x² + 3x = 0. Factoring out -x, we get -x(x - 3) = 0. This implies that x = 0 or x = 3. So, the parabola crosses the x -axis at (0, 0) and (3, 0).

For a parabola in the form y = ax² + bx + c, the x -coordinate of the vertex is given by x = -(b)/(2a). Here, a = -1 and b = 3, so x = -(3)/(2(-1)) = (3)/(2). Substituting this value back into the equation, we find the y -coordinate: y = -((3)/(2))² + 3((3)/(2)) = -(9)/(4) + (9)/(2) = (9)/(4). Thus, the vertex is at ((3)/(2), (9)/(4)).

For a parabola in the form y = ax² + bx + c, the x -coordinate of the vertex is given by x = -(b)/(2a). Here, a = -1 and b = 3, so x = -(3)/(2(-1)) = (3)/(2). Substituting this value back into the equation, we find the y -coordinate: y = -((3)/(2))² + 3((3)/(2)) = -(9)/(4) + (9)/(2) = (9)/(4). Thus, the vertex is at ((3)/(2), (9)/(4)).

what is x square - 3 x + $Y = 0$

Answer: The equation $x^2 - 3x + y = 0$ represents a parabola opening downwards with x-intercepts at $(0, 0)$ and $(3, 0)$ , and its vertex at $(\frac{3}{2}, \frac{9}{4})$ .

Step-by-step solution

Step 1: Identify the equation type

The given equation is $x^2 - 3x + y = 0$ . This equation involves two variables, $x$ and $y$ , and the highest power of $x$ is 2, while the highest power of $y$ is 1. This structure indicates that it represents a parabola.

Step 2: Rearrange the equation to standard form

To better understand the parabola's properties, we can rearrange the equation to express $y$ in terms of $x$ . By moving the $x$ terms to the right side of the equation, we get $y = -x^2 + 3x$ . This is the standard form for a parabola opening downwards.

Step 3: Find the roots (x-intercepts)

To find where the parabola intersects the $x$ -axis, we set $y = 0$ . This gives us $-x^2 + 3x = 0$ . Factoring out $-x$ , we get $-x(x - 3) = 0$ . This implies that $x = 0$ or $x = 3$ . So, the parabola crosses the $x$ -axis at $(0, 0)$ and $(3, 0)$ .

Step 4: Find the vertex

For a parabola in the form $y = ax^2 + bx + c$ , the $x$ -coordinate of the vertex is given by $x = -\frac{b}{2a}$ . Here, $a = -1$ and $b = 3$ , so $x = -\frac{3}{2(-1)} = \frac{3}{2}$ . Substituting this value back into the equation, we find the $y$ -coordinate: $y = -(\frac{3}{2})^2 + 3(\frac{3}{2}) = -\frac{9}{4} + \frac{9}{2} = \frac{9}{4}$ . Thus, the vertex is at $(\frac{3}{2}, \frac{9}{4})$ .

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what is x square - 3 x + Y=0Y = 0Y=0