Question

NO LINKS!! Please help me. Part 2​

Answer 1

We need equation of the parabolas

#1

• Vertex is at (6,4)

One x intercept is present (4,0)

Now put that in vertex form of parabola

• y=a(x-h)²+k
• 0=a(4-6)²+4
• 4a+4=0
• a=-1

So

equation

• y=-(x-6)²+4

#2

• Vertex at (-5,0)

So equation

• y=a(x+5)²+0=a(x+5)²

See the general quadratic equation having vertex (0,0) is shifted to (-5,0) means 5 units left .

So there is no a means a=

Equation remains same

• y=(x+5)²

Answer 2

11)
`y=-(x-6)^2+4`

12)
`y=(x+5)^2`

Explanation:

Vertex form of a parabola:

`y=a(x-h)^2+k` where (h, k) is the vertex

Question 11

From inspection of the graph, the vertex is (6, 4)

`\implies y=a(x-6)^2+4`

To find
`a`, substitute the coordinates of a point on the curve into the equation.

Using point (4, 0):

`\implies a(4-6)^2+4=0`

`\implies a(-2)^2+4=0`

`\implies 4a=-4`

`\implies a=-1`

Therefore, the equation of the parabola in vertex form is:

`y=-(x-6)^2+4`

Question 12

From inspection of the graph, the vertex is (-5, 0)

`\implies y=a(x+5)^2`

To find
`a`, substitute the coordinates of a point on the curve into the equation.

Using point (-3, 4):

`\implies a(-3+5)^2=4`

`\implies a(2)^2=4`

`\implies 4a=4`

`\implies a=1`

Therefore, the equation of the parabola in vertex form is:

`y=(x+5)^2`