Question

I have no idea how to do this stuff.

Answer 1

Answer: $y=2x^{2}-5x+7$

Step-by-step explanation:

The quadratic equation in its standard form is:

$y=ax^{2}+bx+c$

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 $a$, $b$ and $c$ in the quadratic equation.

So, let's choose the first three points:

(-1,14):

$14=a(-1)^{2}+b(-1)+c$

$14=a-b+c$ (1)

(0,7):

$7=a(0)^{2}+b(0)+c$

$7=c$ (2)

(1,4):

$y=2x^{2}-5x+7$

$4=a(1)^{2}+b(1)+c$

$4=a+b+c$ (3)

Substituting (2) in (1) and (3):

$14=a-b+7$ (4)

$4=a+b+7$ (5)

At this point we have a system with two equations.

Adding (4) to (5):

$18=2a+14$ (6)

Isolating $a$:

$a=2$ (7)

Substituitng (7) in (3):

$4=2+b+7$ (8)

Isolating $b$:

$b=-5$ (9)

Now we have the three coeficients and we can write the quadratic equation:

$y=2x^{2}-5x+7$