Answer: [tex]y=2x^{2}-5x+7[/tex]
Step-by-step explanation:
The quadratic equation in its standard form is:
[tex]y=ax^{2}+bx+c[/tex]
Now, we are given 5 points of the parabola (if you graph a quadratic equation you will have a parabola), however we only need to choose three points to find the coeficients [tex]a[/tex], [tex]b[/tex] and [tex]c[/tex] in the quadratic equation.
So, let's choose the first three points:
(-1,14):
[tex]14=a(-1)^{2}+b(-1)+c[/tex]
[tex]14=a-b+c[/tex] (1)
(0,7):
[tex]7=a(0)^{2}+b(0)+c[/tex]
[tex]7=c[/tex] (2)
(1,4):
[tex]y=2x^{2}-5x+7[/tex]
[tex]4=a(1)^{2}+b(1)+c[/tex]
[tex]4=a+b+c[/tex] (3)
Substituting (2) in (1) and (3):
[tex]14=a-b+7[/tex] (4)
[tex]4=a+b+7[/tex] (5)
At this point we have a system with two equations.
Adding (4) to (5):
[tex]18=2a+14[/tex] (6)
Isolating [tex]a[/tex]:
[tex]a=2[/tex] (7)
Substituitng (7) in (3):
[tex]4=2+b+7[/tex] (8)
Isolating [tex]b[/tex]:
[tex]b=-5[/tex] (9)
Now we have the three coeficients and we can write the quadratic equation:
[tex]y=2x^{2}-5x+7[/tex]
