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

Answer:

11)  $y=-(x-6)^2+4$

12)  $y=(x+5)^2$

Step-by-step 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:

$y=(x+5)^2$