Question

Y = 3(x-2)^2 + 5

Answer 1

The y-intercept coordinate is at the point (0, 17)

Properties of functions

Given the following function expressed as:

$y = 3(x-2)^2 + 5$

• The vertex of the function (h, k) is (2, 5) while the axis of symmetry is at x = 2
• Domain are the input values for which the function exist. According to the function, the domain and  range exists on all real values

The x-intercept is the point where y = 0

$3(x-2)^2 + 5=0\\3(x-2)^2 = -5\\(x-2)^2 = -5/3\\x - 2 = 25/9\\x = 25/9 + 2\\x = \frac{43}{9}$

Hence the x-intercept is (43/9, 0)

The y-intercept is the point where x = 0

$y=3(x-2)^2 + 5\\y=3(-2)^2 + 5\\y=12+5\\y=17$

The y-intercept coordinate is at the point (0, 17)

Learn more on vertex and axis of symmetry here: brainly.com/question/21191648