Question

Can someone help me?

Answer 1

(i)
`y = 2x(x-5)`

(ii) zeros: x = 0 and x = 5

axis of symmetry: x = 2.5

(iii) vertex: (2.5, -12.5)

(iv) see attached

Explanation:

Part (i)

Factor:
`y=2x^2-10x`

Factor out the common term
`2x`:

`\implies y=2x(x-5)`

Part (ii)

Zeros occur when
`y=0`

`\implies 2x(x-5)=0`

Therefore:

`\implies 2x=0 \implies x=0`

`\implies (x-5)=0 \implies x=5`

So the zeros are x = 0 and x = 5

The axis of symmetry of a parabola is x = a where a is the midpoint of the zeros.

`\textsf{midpoint of zeros}=(0+5)/(2)=2.5`

Therefore, the axis of symmetry is
`x=2.5`

Part (iii)

The axis of symmetry is the x-coordinate of the vertex. To find the y-coordinate of the vertex, substitute this into the given equation.

`\implies 2(2.5)^2-10(2.5)=-12.5`

So the vertex is (2.5, -12.5)

Part (iv)

Plot the zeros and the vertex.

As the leading coefficient is positive, the parabola will open upwards, so the vertex is the minimum point.

Draw the axis of symmetry, then sketch the parabola, ensuring each side of the axis of symmetry is symmetrical.

*see attached*