Question

Match the equation y=(2x+5)(x−1) with the figure that shows the zeros, vertex, and parabola graphed correctly for the equation.

Answer 1

The given equation is:

$y=(2x+5)(x-1)$

We need to transform it into the form:

$y=ax^2+bx+c$

We can do it by multiplying the parentheses:

$\begin{gathered} y=2x\cdot x-2x\cdot1+5\cdot x-5\cdot1 \\ y=2x^2-2x+5x-5 \\ y=2x^2+3x-5 \end{gathered}$

Then a=2, b=3 and c=-5.

The vertex of the parabola is given by (h,k) where:

$\begin{gathered} h=-\frac{b}{2a} \\ k=f(h) \end{gathered}$

Then h and k are equal to:

$\begin{gathered} h=-\frac{3}{2\cdot2}=-\frac{3}{4}=-0.75 \\ k=f(-0.75) \\ \therefore k=2\cdot(-0.75)^2+3(-0.5)-5 \\ k=1.125-2.25-5 \\ k=-6.13 \end{gathered}$

Thus, the vertex is at (-0.75, -6.13).

Now, to find the zeros of the function, we can use the quadratic formula:

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

By replacing the know values we obtain:

$\begin{gathered} x=\frac{-3\pm\sqrt[]{3^2-4(2)(-5)}}{2(2)} \\ x=\frac{-3\pm\sqrt[]{9+40}}{4} \\ x=\frac{-3\pm\sqrt[]{49}}{4} \\ x=\frac{-3\pm7}{4} \\ \text{Then x is equal to:} \\ x=\frac{-3+7}{4}=\frac{4}{4}=1\text{ and }x=\frac{-3-7}{4}=\frac{-10}{4}=-2.5 \end{gathered}$

Thus, the zeros are at: (-2.5,0) and (1,0).

Answer: the graph that represents the parabola is the last one with the vertex at (-0.75, -6.13) and the zeros at (-2.5,0) and (1,0).