Sketch a graph of y = 4x^2 + 28x + 49 and y = 4x^2 – 28x + 49. Label the vertices.

Question

What can you conclude about the two functions?

Answer 1

The Equations are y = 4x²+28x+49 and y = 4x²-28x+49.

Let us first consider the Equation y = 4x²+28x+49

Now convert this equation in Vertex form of Parabola.

y = 4x²+28x+49

Subtract 49 on both sides.

y - 49 = 4x²+28x

y - 49 = 4(x²+7x)

Now we take term added on both sides is (half of the Coefficient of x)² = (7/2)² = 49/4

y - 49 = 4(x²+7x + 49/4 -49/4)

y - 49 +4(49/4) = 4(x²+7x + 49/4)

y = 4(x - (-7/2))²

Therefore the vertex form of the y = 4x²+28x+49 is y = 4(x - (-7/2))².

Now let us consider the Equation y = 4x² - 28x+49

Now convert this equation in Vertex form of Parabola.

y = 4x² - 28x+49

Subtract 49 on both sides.

y - 49 = 4x² - 28x

y - 49 = 4(x² - 7x)

Now we take term added on both sides is (half of the Coefficient of x)² = (7/2)² = 49/4

y - 49 = 4(x² - 7x + 49/4 -49/4)

y - 49 +4(49/4) = 4(x² - 7x + 49/4)

y = 4(x - 7/2)²

Therefore the vertex form of the y = 4x² - 28x+49 is y = 4(x - 7/2)².

Let us consider two quadratic functions y = 4(x - (-7/2))² and y = 4(x - 7/2)²

y = 4(x - (-7/2))²     and  y = 4(x - 7/2)²

We can observe that the graph is shift from -7/2 to 7/2.

Now graph the functions :

Therefore we can observe there is shift from -3.5 to 3.5 towards right.