Question

Find the zeros of f(x) = (x -3)^2 - 49 graphically

Answer 1

Step-by-step explanation:

First let fully expand this equation

Use the product of squares method.

$(x - y) {}^{2} = {x}^{2} - 2xy + {y}^{2}$

$x {}^{2} - 6x + 9$

Add like terms

${x}^{2} - 6x - 40$

First, let find the vertex.

$- \frac{b}{2a}$

$\frac{6}{2} = 3$

Plug this in to find the minimum point.

${3}^{2} - 6(3) - 40 = - 49$

So the minimum point is at (3,49).

Now we can find some zeroes.

Apply AC Method.

$(x - 10)(x + 4)$

Solve each equation for zero.

$x = 10$

$x = - 4$

So Graph points

(3,49)

(-4,0)

(10,0)