f(x)=(x-1)^2-2 find the vertex, the y intercept and x intercept if any

Question

Pleas show steps.

 

Answer 1

The vertex form of parabola equation is y = a(x - h)^2 + k, where (h, k) = vertex and axis of symmetry x = h.

The parabola is f(x) = (x - 1)² - 2.

Compare the equation (x - 1)² - 2 with the vertex form of parabola equation is y = a(x - h)^2 + k, where (h, k) = vertex and axis of symmetry x = h.

Vertex (h, k) = (1, - 2), and axis of symmetry x = 1.

Find the y -  intercept :

To find the y - intercept, Substitute the value x = 0 in the original function, f(x) = (x - 1)² - 2.

y = f(0) = (0 - 1)² - 2

y = - 1.

The y - intercept is - 1.

Find the x -  intercept :

To find the x - intercept, Substitute the value y = 0 in the original function, f(x) = (x - 1)² - 2.

0 = (x - 1)² - 2

(x - 1)² = 2

(x - 1)² = (± √2)²

x - 1 = ± √2

x = 1 ± √2.

The x - intercepts are 1 - √2 and 1 + √2.