Respuesta :
1. find equation
general solutions are:
[tex]y = a x^{2} +bx +c \\ \\ \frac{dy}{dx} = 2ax +b[/tex]
for P(0,-11)
[tex]-11 = 0a + 0b +c [/tex] ⇒[tex]c = -11[/tex]
for p(1,-13)
[tex]-13 = 1a + 1b -11 \\ a+b = -2[/tex]
and
[tex]0 = 2a + b \\ b = -2a[/tex]
solve for a and b:
[tex]a - 2a = -2 \\ a = 2 \\ b = -4[/tex]
the total equation is now:
[tex]y = 2 x^{2} -4x -11[/tex]
To find the x-intercept set y=0 and solve for x
[tex]0 = 2 x^{2} -4x-11[/tex]
general solutions are:
[tex]y = a x^{2} +bx +c \\ \\ \frac{dy}{dx} = 2ax +b[/tex]
for P(0,-11)
[tex]-11 = 0a + 0b +c [/tex] ⇒[tex]c = -11[/tex]
for p(1,-13)
[tex]-13 = 1a + 1b -11 \\ a+b = -2[/tex]
and
[tex]0 = 2a + b \\ b = -2a[/tex]
solve for a and b:
[tex]a - 2a = -2 \\ a = 2 \\ b = -4[/tex]
the total equation is now:
[tex]y = 2 x^{2} -4x -11[/tex]
To find the x-intercept set y=0 and solve for x
[tex]0 = 2 x^{2} -4x-11[/tex]
Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex
using given coordinates of vertex:
y=A(x-1)^2-13
using given coordinates of y-intercept:
-11=A(0-1)^2-13
A=2
Equation of parabola: y=2(x-1)^2-13
x-intercepts
set y=0
0=2(x-1)^2-13
2(x-1)^2=13
(x-1)^2=13/2
x-1=±√(13/2)
x=1±√(13
using given coordinates of vertex:
y=A(x-1)^2-13
using given coordinates of y-intercept:
-11=A(0-1)^2-13
A=2
Equation of parabola: y=2(x-1)^2-13
x-intercepts
set y=0
0=2(x-1)^2-13
2(x-1)^2=13
(x-1)^2=13/2
x-1=±√(13/2)
x=1±√(13
