Question

Intercepts and the vertex of the graph of the function. Then sketch the graph of the function. $$ y=(x-4)(x+2) $$

Short answer

The intercepts of the function are at x = 4 and x = -2. The vertex is at (3,-1). The graph is a parabola opening upwards through the point (3,-1) with intercepts at x = 4 and x = -2.

Step-by-step solution

Find the intercepts

Intercepts are where the graph crosses or touches the x-axis. To find the intercepts, the equation would be set to 0 and then the x-values would be solved for when y=0. Hence, solving the equation (x-4)(x+2) = 0 yields the two x-intercepts: x = 4 and x = -2.

Find the Vertex

The vertex of a parabola is its highest or lowest point. For a quadratic function in the form y=(x-a)(x+b), the vertex occurs at x=(-b+a)/2. In this case, a = 4 and b = -2. So, the x-coordinate of the vertex is x=(-b+a)/2=(-(-2) + 4)/2 = 3. Substituting x = 3 into the equation gives the y-coordinate of the vertex as y=(3-4)(3+2)=-1. Therefore, the vertex is at (3,-1).

Sketch the graph

To sketch the graph, start by plotting the x-intercepts at x = 4 and x = -2. Next, plot the vertex point at (3,-1). As this is a quadratic function, the graph will be a parabola. It opens upwards, because the leading coefficient (the multiplier of \( x^{2} \)) is positive. Draw a smooth curve that goes through the vertex and the intercepts.