Find the Properties y^2-3y-x+4=0

Problem

y2−3*y−x+4=0

Solution

1. Identify the type of conic section by observing that only one variable, y is squared. This indicates the equation represents a parabola that opens horizontally.

2. Rewrite the equation in standard form (y−k)2=4*p*(x−h) by isolating the y terms and completing the square.

y2−3*y=x−4

3. Complete the square for the y terms by adding ((−3)/2)2=9/4 to both sides of the equation.

y2−3*y+9/4=x−4+9/4

4. Simplify both sides to reach the standard form.

(y−3/2)2=x−7/4

5. Determine the vertex (h,k) from the standard form (y−3/2)2=1*(x−7/4)

Vertex=(7/4,3/2)

6. Calculate the value of p by setting 4*p equal to the coefficient of the x term, which is 1

4*p=1⇒p=1/4

7. Find the focus (h+p,k) since the parabola opens to the right (p>0.

Focus=(7/4+1/4,3/2)=(2,3/2)

8. Identify the directrix using the formula x=h−p

x=7/4−1/4⇒x=3/2

9. Identify the axis of symmetry using the formula y=k

y=3/2

Final Answer

Vertex: *(7/4,3/2), Focus: *(2,3/2), Directrix: *x=3/2, Axis of Symmetry: *y=3/2

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