Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar.

Answer 1

Answer:

$\large\boxed{y=\dfrac{1}{2}x^2-2x+1}$

Step-by-step explanation:

The vertex form of an equation of a parabola:

$y=a(x-h)^2+k$

(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:

$1=a(0-2)^2-1$     add 1 to both sides

$1+1=a(2)^2-1+1$

$2=4a$              divide both sides by 4

$\dfrac{2}{4}=\dfrac{4a}{4}\\\\\dfrac{1}{2}=a\to a=\dfrac{1}{2}$

Therefore we have the equation:

$y=\dfrac{1}{2}(x-2)^2-1$

Convert to the standard form:

$y=\dfrac{1}{2}(x-2)^2-1$       use (a - b)² = a² - 2ab + b²

$y=\dfrac{1}{2}(x^2-2(x)(2)+2^2)-1$

$y=\dfrac{1}{2}(x^2-4x+4)-1$           use the distributive property

$y=\dfrac{1}{2}x^2-\dfrac{1}{2}\cdot4x+\dfrac{1}{2}\cdot4-1$

$y=\dfrac{1}{2}x^2-2x+2-1$          combine like terms

$y=\dfrac{1}{2}x^2-2x+1$