Question

Can i have help in these next few minutes

Answer 1

`y=-(1)/(2)(x-1)^2+2`

Explanation:

Vertex form of a quadratic relation

`y=a(x-h)^2+k`

where:

• (h, k) is the vertex
• a is some constant
  a > 0 : parabola opens upwards
  a < 0 : parabola opens downwards

From inspection of the graph:

• Vertex = (1, 2)
• Point on the curve = (-3, -6)

Substitute the given vertex and point into the vertex formula to find a:

`\implies -6=a(-3-1)^2+2`

`\implies -6=16a+2`

`\implies 16a=-8`

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

Substitute the found value of a and the given vertex into the formula to create the equation for the graph in vertex form:

`y=-(1)/(2)(x-1)^2+2`

Answer 2

• y = - 0.5(x - 1)² + 2

Explanation:

The vertex form is

• y = a(x - h)² + k, where (h, k) is the coordinates of vertex

We see the vertex has coordinates (1, 2)

Substitute the coordinates to get

• y = a(x - 1)² + 2

Use one of the x-intercepts to find the value of a, use (- 1, 0)

• 0 = a(- 1 - 1)² + 2
• 0 = 4a + 2
• 4a = - 2
• a = - 2/4
• a = - 1/2 or - 0.5

The equation of the parabola is

• y = - 0.5(x - 1)² + 2