Answer:
y = [tex]\frac{1}{2}[/tex] x² - 2x + 1
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
From the graph (h, k) = (2, - 1), thus
y = a(x - 2)² - 1
To find a substitute a point on the curve (0, 1) into the equation
1 = 4a - 1 ( add 1 to both sides )
4a = 2 ( divide both sides by 4 )
a = [tex]\frac{2}{4}[/tex] = [tex]\frac{1}{2}[/tex], hence
y = [tex]\frac{1}{2}[/tex](x - 2)² - 1 ← in vertex form
Expand the factor and simplify
y = [tex]\frac{1}{2}[/tex](x² - 4x + 4) - 1
= [tex]\frac{1}{2}[/tex] x² - 2x + 2 - 1
= [tex]\frac{1}{2}[/tex] x² - 2x + 1 ← in standard form
The equation of parabola is:
[tex]y=\dfrac{1}{2}x^2+(-2)x+1[/tex]
We know that the vertex form of the equation of parabola is given by:
[tex]y=a(x-h)^2+k[/tex]
where the vertex of the parabola is (h,k).
Now, from the graph i.e. provided to us we see that the vertex of the parabola is located at (2,-1)
i.e.
(h,k)=(2,-1)
i.e.
h=2 and k= -1
Hence, we have the equation of the parabola as:
[tex]y=a(x-2)^2+(-1)\\\\i.e.\\\\y=a(x-2)^2-1[/tex]
Now, with the help of a passing through point of the parabola we may easily obtain the value of a.
The parabola passes through (0,1)
Hence, on putting x=0 and y=1 we have:
[tex]1=a(0-2)^2-1\\\\i.e.\\\\1=a\times 4-1\\\\i.e.\\\\4a-1=1\\\\i.e.\\\\4a=1+1\\\\i.e.\\\\4a=2\\\\i.e.\\\\a=\dfrac{2}{4}\\\\i.e.\\\\a=\dfrac{1}{2}[/tex]
Hence, the equation of parabola will be:
[tex]y=\dfrac{1}{2}(x-2)^2-1[/tex]
On expanding the square term we have:
[tex]y=\dfrac{1}{2}(x^2+(-2)^2-2\times x\times 2)-1\\\\i.e.\\\\y=\dfrac{1}{2}(x^2+4-4x)-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2+\dfrac{1}{2}\times 4+\dfrac{1}{2}\times (-4x)-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2+2-2x-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2-2x+2-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2-2x+1[/tex]
