Question

1. determine whether each equation in the table represents the graph of the function shown. select yes or no for each equation.

the graph of the function f(x) is a parabola opening downward with vertex and roots as shown in the grid.
the table has three equations:
f(x) = -(x + 3)(x + 5)
f(x) = -(x - 3)(x - 5)
f(x) = -(x - 4)^2 + 1
for each equation, there are yes and no options to select.

Explanation:

Step1: Analyze the graph's roots

The graph intersects the x - axis at \(x = 3\) and \(x=5\). For a quadratic function in factored form \(f(x)=a(x - r_1)(x - r_2)\), the roots are \(r_1\) and \(r_2\). Also, the parabola opens downward, so \(a<0\).

Step2: Check \(f(x)=-(x + 3)(x + 5)\)

The roots of \(f(x)=-(x + 3)(x + 5)\) are \(x=-3\) and \(x = - 5\), which do not match the roots of the graph (\(x = 3\) and \(x = 5\)). So the answer for this equation is No.

Step3: Check \(f(x)=-(x - 3)(x - 5)\)

The roots of \(f(x)=-(x - 3)(x - 5)\) are \(x = 3\) and \(x=5\), and the coefficient of the quadratic term is \(- 1<0\) (so it opens downward), which matches the graph. So the answer for this equation is Yes.

Step4: Check the vertex form \(f(x)=-(x - 4)^2+1\)

The vertex form of a parabola is \(f(x)=a(x - h)^2+k\), where \((h,k)\) is the vertex. For \(f(x)=-(x - 4)^2+1\), the vertex is \((4,1)\). From the graph, the vertex is at the mid - point of \(x = 3\) and \(x = 5\) (since the axis of symmetry of a parabola with roots \(x_1\) and \(x_2\) is \(x=\frac{x_1 + x_2}{2}=\frac{3 + 5}{2}=4\)) and the y - coordinate of the vertex: when \(x = 4\), let's check the factored form \(f(x)=-(x - 3)(x - 5)=-(4 - 3)(4 - 5)=- (1)\times(-1)=1\). So the vertex is \((4,1)\) and the parabola opens downward (\(a=-1<0\)), which matches the graph. So the answer for this equation is Yes.

Answer:

• For \(f(x)=-(x + 3)(x + 5)\): No
• For \(f(x)=-(x - 3)(x - 5)\): Yes
• For \(f(x)=-(x - 4)^2+1\): Yes

(In the table, for \(f(x)=-(x + 3)(x + 5)\) select No; for \(f(x)=-(x - 3)(x - 5)\) select Yes; for \(f(x)=-(x - 4)^2+1\) select Yes)