Answer:
[tex]\textsf{33.} \quad y=-x^2-2x+3[/tex]
[tex]\textsf{34.} \quad y = x^2+4x+3[/tex]
Step-by-step explanation:
Question 33
Factored form of a quadratic equation:
[tex]\boxed{y=a(x-r_1)(x-r_2)}[/tex]
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 [tex]a[/tex]:
[tex]\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\\\dfrac{-4a}{-4} & = \dfrac{4}{-4}\\a & = -1\end{aligned}[/tex]
Substitute the found value of [tex]a[/tex] and the roots into the formula and rewrite in standard form:
[tex]\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}[/tex]
Therefore, the equation of the parabola in standard form is:
[tex]y=-x^2-2x+3[/tex]
Question 34
Vertex form of a quadratic equation:
[tex]\boxed{y=a(x-h)^2+k}[/tex]
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 [tex]a[/tex]:
[tex]\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\\\dfrac{4a}{4} & = \dfrac{4}{4}\\a & = 1\end{aligned}[/tex]
Substitute the found value of [tex]a[/tex] and the vertex into the formula and rewrite in standard form:
[tex]\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}[/tex]
Therefore, the equation of the parabola in standard form is:
[tex]y & = x^2+4x+3[/tex]
