Respuesta :
A parabola is the shape of the graph of a quadratic function. Quadratic functions have the general form:
[tex]y=ax^2+bx+c[/tex]
Here, you've been given the x and y values for three points on the parabola, and you've been asked to set up a system of equations that include those three points. All this problem is really asking is to plug each of those values into the general equation.
The system should look like:
[tex]\begin{array}{lcl} 3^2a+3b+c & = & 8 \\ 5^2a+5b+c & = & 20/3 \\ 6^2a+6b+c & = & 5 \end{array}[/tex]
Which, when simplified, looks like:
[tex]\begin{array}{lcl} 9a+3b+c & = & 8 \\ 25a+5b+c & = & 20/3 \\ 36a+6b+c & = & 5 \end{array}[/tex]
[tex]y=ax^2+bx+c[/tex]
Here, you've been given the x and y values for three points on the parabola, and you've been asked to set up a system of equations that include those three points. All this problem is really asking is to plug each of those values into the general equation.
The system should look like:
[tex]\begin{array}{lcl} 3^2a+3b+c & = & 8 \\ 5^2a+5b+c & = & 20/3 \\ 6^2a+6b+c & = & 5 \end{array}[/tex]
Which, when simplified, looks like:
[tex]\begin{array}{lcl} 9a+3b+c & = & 8 \\ 25a+5b+c & = & 20/3 \\ 36a+6b+c & = & 5 \end{array}[/tex]
To solve this question:
- First, we need to know the standard equation of a parabola.
- Then, with the three points given, we can build a system of three equations.
Doing this, we get that the system is, with the 3 equations:
[tex]9a^2 + 3b + c = 8[/tex]
[tex]75a + 15b + 3c = 20[/tex]
[tex]36a + 6b + c = 5[/tex]
Equation of a parabola:
A parabola is a second degree polynomial, having an standard equation of:
[tex]y = ax^2 + bx + c, a \neq 0[/tex]
Point (3,8)
This means that when [tex]x = 3, y = 8[/tex]. So
[tex]y = ax^2 + bx + c, a \neq 0[/tex]
[tex]8 = a\times3^2 + 3b + c[/tex]
[tex]9a^2 + 3b + c = 8[/tex]
Thus, [tex]9a^2 + 3b + c = 8[/tex] is the first equation.
(5,20/3)
This means that when [tex]x = 5, y = \frac{20}{3}[/tex]. Thus
[tex]y = ax^2 + bx + c, a \neq 0[/tex]
[tex]\frac{20}{3} = a\times5^2 + 5b + c[/tex]
[tex]25a + 5b + c = \frac{20}{3}[/tex]
[tex]75a + 15b + 3c = 20[/tex]
Thus, [tex]75a + 15b + 3c = 20[/tex] is the second equation.
(6,5)
This means that when [tex]x = 6, y = 5[/tex]. Thus
[tex]y = ax^2 + bx + c, a \neq 0[/tex]
[tex]5 = a\times6^2 + 6b + c[/tex]
[tex]36a + 6b + c = 5[/tex]
Thus, [tex]36a + 6b + c = 5[/tex] is the third equation.
After all, the system of equations is:
[tex]9a^2 + 3b + c = 8[/tex]
[tex]75a + 15b + 3c = 20[/tex]
[tex]36a + 6b + c = 5[/tex]
A similar question is found at https://brainly.com/question/1440592
