Question

Use this graph of y = 2x2 - 12x + 19 to find the vertex. Decide whether thevertex is a maximum or a minimum point.A. Vertex is a minimum point at (3, 1)B. Vertex is a maximum point at (1,7)C. Vertex is a minimum point at (1,3)D. Vertex is a maximum point at (3,1)

Answer 1

Hello there. To slve this question, we'll have to remembrer some properties about maximum and minimum in a quadratic function.

Given a quadratic function f as follows:

`f(x)=ax^2+bx+c`

We can determine whether or not the vertex is a maximum or minimum by the signal of the leading coefficient a.

If a < 0, the concavity ofthe parabolai is facing down, hence it admits a maximum value at its vertex.

If a > 0, the concavity of the parabola is facing up, hence it admits a minimum value at its vertex.

As a cannot be equal to zero (otherwise we wouldn't have a quadratic equation), we use the coefficients to determine an expression for the coordinates of the vertex.

The vertex is, more generally, located in between the roots of the function.

t is easy to prove, y comlpleting hthe square, that the solutions of the equation

`ax^2+bx+c=0`

are given as

`x=(-b\pm√(b^2-4ac))/(2a)`

Taking the arithmetic mean of these values, we get the x-coordinate of the vertex:

`x_V=((-b+√(b^2-4ac))/(2a)+(-b-√(b^2-4ac))/(2a))/(2)=(-(2b)/(2a))/(2)=-(b)/(2a)`

By evaluating the function at this point, we'll obtain the y-coordinate of the vertex:

`f(x_V)=-(b^2-4ac)/(4a)`

With this, we can solve this question.

Given the function:

`y=2x^2-12x+19`

First, notice the leadin coefficient is a = 2 , that is positive.

Hence it has a minimum point at its vertex.

To determine these coordinates, we use the other coefficients b = -12 and c = 19.

Plugging the values, we'll get

`x_V=-(-12)/(2\cdot2)=(12)/(4)=3`

Plugging ths value in the function, ewe'll get

`y_V=f(x_V)=2\cdot3^2-12\cdot3+19=2\cdot9-36+19=1`

Hence we say that the final answer is

Vertex is a minimum point at (3, 1)

As you can see in the gaph.