Question

NO LINKS!! PART 3. Please help me with these graphs​

Answer 1

#1

• Vertex at (-5,4)

Equation

• y=a(x+5)²+4

As there is no y inetercept present and the function is same as general quadratic equation y=x^2 just change in coordinates a=1

Final equation

• y=(x+5)²+4

#2

• Vertex (2,-3)

y=a(x-2)²-3

Use (4,-5)

• -4=a(4-2)²-3
• -1=a(2)²
• 4a=-1
• a=-1/4

Equation

• y=-1/4(x-2)²-3

Answer 2

`\textsf{13)} \quad y=(x+5)^2+4`

`\textsf{14)} \quad y=-(1)/(2)(x-2)^2-3`

Explanation:

Vertex form of a parabola:

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

Question 13

From inspection of the graph, the vertex is (-5, 4)

`\implies y=a(x+5)^2+4`

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

Using point (-4, 5):

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

`\implies a(1)^2+4=5`

`\implies a+4=5`

`\implies a=1`

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

`y=(x+5)^2+4`

Question 14

From inspection of the graph, the vertex is (2, -3)

`\implies y=a(x-2)^2-3`

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

Using point (0, -5):

`\implies a(0-2)^2-3=-5`

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

`\implies 4a-3=-5`

`\implies 4a=-2`

`\implies a=-(1)/(2)`

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

`\implies y=-(1)/(2)(x-2)^2-3`