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In this lesson, we will learn how to identify the properties of the parabola using graphical representation and the function rule.

Q1:

Find the coordinates of the vertex of the function π ( π₯ ) = β 8 π₯ β 6 + 4 π₯ 2 .

Q2:

The figure shows the curve ( π¦ + π₯ β 3 ) + 4 ( π¦ β π₯ ) + 4 = 0 2 , the dashed line π¦ + π₯ = 3 , and its perpendicular π¦ β π₯ = β 5 .

Determine the coordinates of the two intersections π΄ and π΅ .

The vertex of this parabola lies where the line π¦ β π₯ = πΎ meets the parabola at exactly one point.

What is the value of πΎ ? What are the coordinates of the vertex?

The vertex also lies on the line of symmetry π¦ + π₯ = 3 , which is present in the equation. It turns out that the equation ( π¦ + π₯ β 3 ) β 8 ( π¦ + π₯ β 3 ) + 4 ( π¦ β π₯ ) + 2 = 0 2 gives a parabola whose axis of symmetry is parallel to π¦ + π₯ = 3 .

By completing the square and rewriting this as ( π¦ + π₯ β π΄ ) + 4 ( π¦ β π₯ ) + π΅ = 0 2 , determine what the new axis of symmetry is. What is the value of the constant π΅ ?

By determining π΄ , π΅ such that π¦ β 5 π₯ = π΄ ( π¦ + π₯ ) + π΅ ( π¦ β π₯ ) , complete the squares and rewrite ( π¦ + π₯ ) + π¦ β 5 π₯ + 7 2 as ( π¦ + π₯ β π ) + π ( π¦ β π₯ ) + π 2 . What is this expression? What is the vertex of the parabola?

Q3:

A parabola has an equation ( π₯ β 4 ) = 2 0 ( π¦ + 3 ) ο¨ .

Find the coordinates of the vertex.

Determine the equation of the directrix.

Q4:

Consider the parabola with Cartesian equation π¦ = β 2 β 1 7 π₯ ο¨ .

What are the coordinates of its focus?

Write the equation of its directrix.

Q5:

Find the vertex of the graph of π¦ = ( 3 π₯ + 1 ) 2 .

Q6:

Consider the parabola with the Cartesian equation π₯ = β 6 β 1 1 π¦ 2 .

What are the coordinates of the focus of the parabola with the Cartesian equation π₯ = β 6 β 1 1 π¦ 2 ?

Q7:

Find the vertex of the graph of π¦ = ( π₯ β 1 ) 2 .

Q8:

Find the point of symmetry of the curve of the function π ( π₯ ) = β 7 β ( π₯ β 2 ) 2 .

Q9:

Find the vertex of the graph of π¦ = β π₯ β 4 2 .

Q10:

Consider the curve shown below.

Which suitable triple π΄ , π΅ , πΆ would make this the graph of π ( π₯ ) = π΄ ( π₯ β π΅ ) + πΆ 2 ?

Q11:

Find the vertex of the graph of π¦ = β ( π₯ + 2 ) 2 .

Q12:

Find the vertex of the quadratic equation π¦ = β 2 π₯ + 1 2 π₯ β 1 2 .

Q13:

Suppose that π ( π₯ ) = π΄ ( π₯ β 1 ) ( π₯ β 4 ) for some constant π΄ . What is the π¦ -coordinate of the vertex of parabola π¦ = π ( π₯ ) ?

Q14:

A parabola has the equation π¦ = 7 π₯ 2 .

Write an equation for its directrix.

Q15:

Write an equation for the parabola whose focus is the point οΏ β β 5 6 , 0 ο and whose directrix is the line π₯ = β 5 6 .

Q16:

Consider a parabola described by the parametric equations π₯ = 1 1 π‘ 2 , π¦ = 2 2 π‘ , π‘ β β .

What are the coordinates of the focus?

What is the equation of the directrix?

Q17:

The figure shows the curve ( π¦ + π₯ β 3 ) + 4 ( π¦ β π₯ ) + 4 = 0 ο¨ , the dashed line π¦ + π₯ = 3 , and its perpendicular π¦ β π₯ = β 5 .

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