Question

Which of the following equations fits this graph?

Answer 1

Answer: option A

Step-by-step explanation:

You can observe in the graph that one point of the parabola is (2,1)

Then, substittute x=2 into each equation:

$A.\ y=-2x^2+4x+1\\y=-2(2)^2+4(2)+1\\y=1$

The point obtained is (2,1), which belong to the parabola.

$B.\ y=2x^2+4x+1\\y=2(2)^2+4(2)+1\\y=17$

The point obtained is (2,17), which does not belong to the parabola.

$C.\ y=-2x^2+x+1\\y=2(2)^2+4(2)+1\\y=-5$

The point obtained is (2,-5), which does not belong to the parabola.

$D.\ y=2x^2+2x-1\\y=2(2)^2+4(2)+1\\y=11$

The point obtained is (2,11), which does not belong to the parabola.

Answer 2

Hello!

The answer is:

The equation A. $y=-2x^{2} +4x+1$ has its vertex on the point (1,3) fits with the graph.

Why?

The given parabola has a vertex on the point (1,3) and cuts the y-axis at y equal to 1, also, we know that it cuts the x-axis at number between -1 and 0, and a number between 2 and 3, however, the simplest way to know which if the given equations fits with the graph. we need to look for the equation of the parabola that has a vertex on the point (1,3).

The vertex of the parabola is the highest or lowest point that the parabola has, depending on if it's opening downwards or upwards.

We need that the standard form of the parabola is:

$y=ax^{2} +bx+c$

and, we can find the vertex of the parabola using the following equation:

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

Then, we need to substitute the "x" value into the equation of the parabola to find the y-coordinate value.

Also, when the coefficient of the quadratic term (a) is negative, the parabola opens downwards, on the opposite, when it's positive, the parabola opens upwards.

To solve find which of the following equations fits this graph, since we can see that the parabola on the graph is opening downwards, we need to look only for the equations that has a quadratic term with a negative coefficients, so, the equation B and D are discarded since both have positive quadratic terms, and they open upwards.

So, looking for the vertex that is located at (1,3) for the equation A, and C, we have:

- First equation, A. $y=-2x^{2} +4x+1$

Where,

$a=-2\\b=4\\c=1$

Finding the vertex, we have:

$x=\frac{-b}{2a}=\frac{-4}{2*(-2)}=\frac{-4}{-4}=1$

The, substituting "x" into the parabola equation, we have:

$y=-2x^{2} +4x+1\\\\y=-2(1)^{2}+4*(1)+1=-2+4+1=3$

Then, we have that the parabola has its vertex on the point (1,3).

- Third equation, C. $y=-2x^{2} +x+1$

$a=-2\\b=1\\c=1$

$x=\frac{-b}{2a}=\frac{-1}{2*(-2)}=\frac{-1}{-4}=0.25$

The, substituting "x" into the parabola equation, we have:

$y=-2x^{2} +4x+1\\\\y=-2(0.25)^{2}+(0.25)+1=-2*(0.063)+0.25+1=1.125$

Then, we have that the parabola has its vertex on the point (0.25,1.125).

Hence, the equation A. $y=-2x^{2} +4x+1$ has its vertex on the point (1,3), meaning that it fits with the graph.

Have a nice day!