Answer:
The vertex happens at ([tex]\frac{4}{5} , \frac{4}{5}[/tex])
Step-by-step explanation:
First of all, we know that the y - intercept is 4, this gives us a third point which is (0,4)
We know that the equation of the parabola: y = [tex]Ax^{2} + Bx + c[/tex].
We also have three points, this means we will end up with three equations and therefore, we would need to solve for A, B and C.
First equation: 4 = A(0) +B(0) + C , C= 4
Second equation: 8 = A(2)^2 + B(2) + 4,
8 = 4A + 2B + 4
4 = 4A + 2 B (Now we need to divide by 4)
1 = A + [tex]\frac{1}{2}[/tex]B where A = 1 - 0.5B
Third equation: 1 = A(1) + B(1) + 4, we know A = 1 -0.5 B so we can substitute it and solve for B.
-3 =1 -0.5(b) + B, We get b = -8
We also figured out that A = 1 - 0.5B and b = -8, therefore A = 5.
Therefore, the equation of the parabola is y = [tex]5x^{2} - 8x + 4[/tex]
In order to find the vertex, we take the derivative and equate it to 0.
y' = 10x -8= 0,
x = [tex]\frac{4}{5}[/tex]
Now we now the x-coordinate of the vertex, in order to find the y-coordinate we have to substitute it into the equation.
y = [tex]5(\frac{4}{5})^{2} - 8 \frac{4}{5} + 4[/tex],
therefore, y = [tex]\frac{4}{5}[/tex].
Therefore, the vertex is at the point [tex](\frac{4}{5}, \frac{4}{5} )[/tex]
Great question, haven't done these since highschool had to look up some stuff.
Hope this helps.
