Question

Can someone help me with this math hw ​

Answer 1

Explanation:

F(x) = x² - 2x + 1

= (x - 1)²

By comparing this equation with the vertex form of the quadratic equation,

y = (x - h)² + k

Here, (h, k) is the vertex

Vertex of the parabola → (1, 0)

x-intercepts → (x - 1)² = 0

x = 1

y-intercepts → y = (0 - 1)²

y = 1

Now we can draw the graph of the given function,

From this graph,

As x → 0,

`\lim_(x \to 0^(-)) (x-1)^2=1`

`\lim_(x \to 0^(+)) (x-1)^2=1`

f(0) = (0 - 1)²

= 1

Since,
`\lim_(x \to 0^(-)) (x-1)^2=\lim_(x \to 0^(+)) (x-1)^2=1`

Therefore, given function is continuous at x = 0.