Answer:
[tex]\large\boxed{y=\dfrac{1}{2}x^2-2x+1}[/tex]
Step-by-step explanation:
The vertex form of an equation of a parabola:
[tex]y=a(x-h)^2+k[/tex]
(h, k) - vertex
a - leading coefficient in equation y = ax² + bx + c
From the grap we can read coordinates of the vertex (2, -1) and y-intercept (0, 1).
Therefore h = 2, k = -1
Put the values of h, k and coordinates of the y-intercept to the equation of parabola:
[tex]1=a(0-2)^2-1[/tex] add 1 to both sides
[tex]1+1=a(2)^2-1+1[/tex]
[tex]2=4a[/tex] divide both sides by 4
[tex]\dfrac{2}{4}=\dfrac{4a}{4}\\\\\dfrac{1}{2}=a\to a=\dfrac{1}{2}[/tex]
Therefore we have the equation:
[tex]y=\dfrac{1}{2}(x-2)^2-1[/tex]
Convert to the standard form:
[tex]y=\dfrac{1}{2}(x-2)^2-1[/tex] use (a - b)² = a² - 2ab + b²
[tex]y=\dfrac{1}{2}(x^2-2(x)(2)+2^2)-1[/tex]
[tex]y=\dfrac{1}{2}(x^2-4x+4)-1[/tex] use the distributive property
[tex]y=\dfrac{1}{2}x^2-\dfrac{1}{2}\cdot4x+\dfrac{1}{2}\cdot4-1[/tex]
[tex]y=\dfrac{1}{2}x^2-2x+2-1[/tex] combine like terms
[tex]y=\dfrac{1}{2}x^2-2x+1[/tex]
