Respuesta :
1.
A "parabolic" function, is also called a "Quadratic" function.
Any quadratic function can be written in the form:
[tex]f(x)=ax^2+bx+c, [/tex] where a≠0, generally called the "standard form".
The set of all points (x, f(x)) plotted in a coordinate axes, forms a parabola.
2.
The parabola contains the points
A(-5, 0), B(-4, -1), C(-3, 0), D(0, 15)
from the above discussion we have:
A(-5, 0) = A(-5, f(-5)),
so f(-5)=0
[tex]f(-5)=a(-5)^2+b(-5)+c\\\\0=25a-5b+c[/tex],
similarly:
[tex]f(-4)=a(-4)^2+b(-4)+c\\\\-1=16a-4b+c[/tex]
and
[tex]f(0)=a(0)^2+b(0)+c\\\\15=c[/tex]
the last equation is particularly important, because it tells us that c=15.
3.
using the first 2 equations we write the system of linear equations:
[tex]\begin{cases} i) 25a-5b+15=0\\ii) 16a-4b+15=-1 \end{cases}\\\\\\\\\begin{cases} i) 25a-5b=-15\\ii) 16a-4b=-16 \end{cases}[/tex]
divide the first equation by 5, and the second one by -4:
[tex]\begin{cases} i) 5a-b=-3\\ii) -4a+b=4 \end{cases}[/tex]
add the 2 equations:
a=1,
then, substituting in any of the equations:
-4a+b=4
-4+b=4
b=8,
4.
thus the function is [tex]f(x)=x^2+8x+15[/tex]
Remark:
given any 3 points of a parabola, it is possible to write the quadratic function.
we did not use 4 of the points, 3 were enough.
A "parabolic" function, is also called a "Quadratic" function.
Any quadratic function can be written in the form:
[tex]f(x)=ax^2+bx+c, [/tex] where a≠0, generally called the "standard form".
The set of all points (x, f(x)) plotted in a coordinate axes, forms a parabola.
2.
The parabola contains the points
A(-5, 0), B(-4, -1), C(-3, 0), D(0, 15)
from the above discussion we have:
A(-5, 0) = A(-5, f(-5)),
so f(-5)=0
[tex]f(-5)=a(-5)^2+b(-5)+c\\\\0=25a-5b+c[/tex],
similarly:
[tex]f(-4)=a(-4)^2+b(-4)+c\\\\-1=16a-4b+c[/tex]
and
[tex]f(0)=a(0)^2+b(0)+c\\\\15=c[/tex]
the last equation is particularly important, because it tells us that c=15.
3.
using the first 2 equations we write the system of linear equations:
[tex]\begin{cases} i) 25a-5b+15=0\\ii) 16a-4b+15=-1 \end{cases}\\\\\\\\\begin{cases} i) 25a-5b=-15\\ii) 16a-4b=-16 \end{cases}[/tex]
divide the first equation by 5, and the second one by -4:
[tex]\begin{cases} i) 5a-b=-3\\ii) -4a+b=4 \end{cases}[/tex]
add the 2 equations:
a=1,
then, substituting in any of the equations:
-4a+b=4
-4+b=4
b=8,
4.
thus the function is [tex]f(x)=x^2+8x+15[/tex]
Remark:
given any 3 points of a parabola, it is possible to write the quadratic function.
we did not use 4 of the points, 3 were enough.
Solution: The correct option is second option, i.e., [tex]y=(x+4)^2-1[/tex].
Explanation:
The standard form of the parabola along the y-axis with vertex (h,k) is [tex]y=a(x-h)^2+k[/tex].
Since the turning point is given as [tex](-4,-1)[/tex].
Put these values in the standard form of the parabola.
[tex]y=a(x+(-4))^2+(-1)[/tex]
[tex]y=a(x+4)^2-1[/tex] .....(1)
The parabola passes through the points (-5,0), (-3,0) and (0,15). It means each point will satisfy the above condition.
Put x = 0 and y = 15 in the equation (1).
[tex]15=a(0+4)^2-1\\16=4^2a\\16=16a\\a=1[/tex]
Put a = 1 in equation (1).
[tex]y=1(x+4)^2-1[/tex]
Therefore, the The correct option is second option, i.e., [tex]y=(x+4)^2-1[/tex].
