graph the equation ( y = -x^{2}-14x - 48 ) on the accompanying set of axes. you must plot 5 points including…

graph the equation ( y = -x^{2}-14x - 48 ) on the accompanying set of axes. you must plot 5 points including the roots and the vertex. using the graph, determine the vertex of the parabola. click to plot points. click points to delete them.

graph the equation ( y = -x^{2}-14x - 48 ) on the accompanying set of axes. you must plot 5 points including the roots and the vertex. using the graph, determine the vertex of the parabola. click to plot points. click points to delete them.

Answer

Explanation:

Step1: Find the roots

Set (y = 0), so (-x^{2}-14x - 48=0), multiply by (- 1) gives (x^{2}+14x + 48 = 0). Factor: ((x + 6)(x+8)=0). Solutions: (x=-6) and (x = - 8). So the points are ((-6,0)) and ((-8,0)).

Step2: Find the vertex

For a quadratic function (y = ax^{2}+bx + c), the (x) - coordinate of the vertex is (x=-\frac{b}{2a}). Here (a=-1), (b=-14), so (x =-\frac{-14}{2\times(-1)}=-7). Substitute (x = - 7) into (y=-x^{2}-14x - 48): (y=-(-7)^{2}-14\times(-7)-48=-49 + 98-48=1). The vertex is ((-7,1)).

Step3: Find two more points

Let (x=-5), then (y=-(-5)^{2}-14\times(-5)-48=-25 + 70-48=-3). So the point is ((-5,-3)). Let (x=-9), then (y=-(-9)^{2}-14\times(-9)-48=-81+126 - 48=-3). So the point is ((-9,-3)).

Answer:

The vertex of the parabola (y=-x^{2}-14x - 48) is ((-7,1)).