Since we know both of the parabola's x-intercepts, we can write the function in factored form.
B
b
Since we know both of the parabola's x-intercepts, we can write the function in factored form.
C
c
Note that this parabola has a double root.
A
a
y=-0.25(x+9)(x-5)
B
b
y=2(x+7)(x+4)
C
c
y=1/3(x+6)^2
Practice makes perfect
a
From the graph we can identify both of the parabola's x-intercepts, x_1=-9 and x_2=5, as well as its y-intercept, y=11.25. Since we have been given both x-intercepts, we can write the function in factored form.
y=a(x-x_1)(x-x_2)
In this form the x-intercepts lie at (x_1,0) and (x_2,0). Therefore, we should substitute x_1= -9 and x_2= 5 into the factored form.
y=a(x-( -9))(x- 5)
⇕
y=a(x+9)(x-5)
To find the value of a we have to substitute a third point in our equation that is not the x-intercepts. Luckily, we have been given the y-intercept as previously mentioned.
Now we can complete the equation.
y=-0.25(x+9)(x-5)
b
Like in Part A, we have been given the parabolas x-intercepts, x_1=-7 and x_2=- 4, as well as its vertex. By substituting x_1= -7 and x_2= -4 into this factored form of a parabola, we are on our way to writing our function.
y=a(x-( -7))(x-( - 4))
⇕
y=a(x+7)(x+4)To find the value of a we substitute the vertex into the equation and solve for a.
In this case we have a double root at x=-6. Therefore, x_1= x_2= 6, which gives us the following factored form.
y=a(x-( - 6))(x-( - 6))
⇕
y=a(x+6)(x+6)Like in Part A, we have been given the y-intercept. By substituting this into the equation we can solve for a.
