Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

We've looked at how to solve and graph quadratics, as well as how to find the concavity, axis of symmetry, turning points and intercepts of parabolas.

Summary

- Concavity: if the coefficient of $x^2$
`x`2 is positive, it's concave up. If the coefficient is negative, it's concave down. - Axis of symmetry: use the formula $x=\frac{-b}{2a}$
`x`=−`b`2`a`. - Turning point: Substitute the equation of the axis of symmetry into the original equation.
- Intercepts: Substitute $y=0$
`y`=0 and $x=0$`x`=0 to find the $x$`x`- and $y$`y`-intercepts, respectively.

Now we are going to apply this knowledge and use it to solve real-life problems.

QUESTION 1

A rectangle is to be constructed with $80$80 metres of wire. The rectangle will have an area of $A=40x-x^2$`A`=40`x`−`x`2, where $x$`x` is the length of one side of the rectangle.

Using the equation, state the area of the rectangle if one side is $12$12 metres long.

The graph below displays all the possible areas that can be obtained using this amount of wire. From the graph, determine the nearest value for the longer side of a rectangle that has an area of $256$256 square metres.

Loading Graph...$33$33 m

A$9$9 m

B$8$8 m

C$32$32 m

DUsing the graph, what is the greatest possible area of a rectangle that has a perimeter of $80$80 m?

Using the graph, state the dimensions of the rectangle with the maximum area.

Length $=$=$\editable{}$ m

Width $=$=$\editable{}$ m

On the moon, the equation $d=0.8t^2$`d`=0.8`t`2 is used to approximate the distance an object has fallen after $t$`t` seconds. (Assuming no air resistance or buoyancy). On Earth, the equation is $d=4.9t^2$`d`=4.9`t`2.

A rock is thrown from a height of $60$60 metres on Earth. Solve for $t$

`t`, the time it will take to hit the ground.Give your answer to the nearest second.

If a rock is thrown from a height of $60$60 metres on the moon, solve for the time, $t$

`t`, it will take to reach the ground.Give your answer to the nearest second.

In a game of tennis, the ball is mistimed and hit high up into the air. Initially (ie at $t=0$`t`=0), the ball is struck $3.5$3.5 metres above the ground and hits the ground $7$7 seconds later. It reaches its greatest height $3$3 seconds after being hit.

Determine the coordinates of the points labelled A, B and C in the diagram.

A $\left(\editable{},\editable{}\right)$(,)

B $\left(\editable{},\editable{}\right)$(,)

C $\left(\editable{},\editable{}\right)$(,)

Using the form of the parabola $y=a\left(t-h\right)^2+k$

`y`=`a`(`t`−`h`)2+`k`, where $t$`t`is the number of seconds after the ball is hit and $y$`y`is the height of the ball above the ground, determine the value of $h$`h`.By substituting $t=0$

`t`=0 and $y=3.5$`y`=3.5 into the equation, form an algebraic relationship between $a$`a`and $k$`k`.By substituting $t=7$

`t`=7 and $y=0$`y`=0 into the equation, form another algebraic relationship between $a$`a`and $k$`k`.Solve for the exact value of $a$

`a`.Hence find the exact value of $k$

`k`, the greatest height reached by the ball.