Question

NO LINKS!! Please help me with these problems. Part 8a1​

Answer 1

`\textsf{33.} \quad y=-x^2-2x+3`

`\textsf{34.} \quad y = x^2+4x+3`

Explanation:

Question 33

Factored form of a quadratic equation:

`\boxed{y=a(x-r_1)(x-r_2)}`

where:

• a is the leading coefficient.
• r₁ and r₂ are the roots.

From inspection of the graph:

• r₁ = -3
• r₂ = 1
• vertex = (-1, 4)

Substitute the roots and the vertex into the formula and solve for
`a`:

`\begin{aligned}y & =a(x-r_1)(x-r_2)\\\implies 4 & = a(-1-(-3))(-1-1)\\4 & = a(2)(-2)\\4 & = -4a\\-4a & = 4\\(-4a)/(-4) & = (4)/(-4)\\a & = -1\end{aligned}`

Substitute the found value of
`a` and the roots into the formula and rewrite in standard form:

`\begin{aligned}y & =a(x-r_1)(x-r_2)\\\implies y & = -1(x-(-3))(x-1)\\y & = -(x+3)(x-1)\\y & = -(x^2+2x-3)\\y & = -x^2-2x+3\end{aligned}`

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

`y=-x^2-2x+3`

Question 34

Vertex form of a quadratic equation:

`\boxed{y=a(x-h)^2+k}`

where:

• (h, k) is the vertex.
• a is some constant.

From inspection of the graph:

• Vertex = (-2, -1) ⇒ h = -2 and k = -1
• y-intercept = (0, 3)

Substitute the vertex and y-intercept into the formula and solve for
`a`:

`\begin{aligned}y & = a(x-h)^2+k\\\implies 3 & = a(0-(-2))^2+(-1)\\3 & = a(0+2)^2-1\\3 & = 4a-1\\3 +1& = 4a-1+1\\4 & = 4a\\4a & = 4\\(4a)/(4) & = (4)/(4)\\a & = 1\end{aligned}`

Substitute the found value of
`a` and the vertex into the formula and rewrite in standard form:

`\begin{aligned}y & = a(x-h)^2+k\\\implies y & = 1(x-(-2))^2+(-1)\\y & = (x+2)^2-1\\y & = (x+2)(x+2)-1\\y & = x^2+4x+4-1\\y & = x^2+4x+3\end{aligned}`

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

`y & = x^2+4x+3`

Answer 2

33. y = -x² - 2x + 3

34. y = x² + 4x + 3

Equation of a parabola is given by y = a(x - h)² + k

where:

• (h, k) is the vertex

33.

Looking at the graph find the vertex, vertex is at (-1, 4)

Take one point: (1, 0)

Substitute (h, k) = (-1, 4) and (x, y) = (1, 0) into equation to find value of a:

• 0 = a(1 - (-1))² + 4

• 0 = a(2)² + 4

• 4a + 4 = 0

• 4a = -4

• a = -1

Now equation:

y = a(x - h)² + k

y = -1(x - (-1))² + 4

y = -(x + 1)² + 4

y = -(x² + 2x + 1) + 4

y = -x² - 2x - 1 + 4

y = -x² - 2x + 3

34.

The vertex is at (h, k) = (-2, -1) and the curve passes through (x, y) = (0, 3)

Find value of a:

• y = a(x - h)² + k

• 3 = a(0 - (-2))² - 1

• 3 + 1 = a(2)²

• 4a = 4

• a = 1

Now equation:

y = a(x - h)² + k

y = 1(x - (-2))² - 1

y = 1(x + 2)² - 1

y = (x² + 4x + 4) - 1

y = x² + 4x + 3