Question

A quadratic equation is shown below:

Answer 1

Answer:

Part A:

( 1.8333, -0.08333)

Part B:

x = 2 or x = 5/3

Step-by-step explanation:

The quadratic equation

$3x^{2}-11x+10=0$ has been given.

Part A:

We are required to determine the vertex. The vertex is simply the turning point of the quadratic function. We shall differentiate the given quadratic function and set the result to 0 in order to obtain the co-ordinates of its vertex.

$\frac{d}{dx}(3x^{2}-11x+10)=6x-11$

Setting the derivative to 0;

6x - 11 = 0

6x = 11

x = 11/6

The corresponding y value is determined by substituting x = 11/6 into the original equation;

y = 3(11/6)^2 - 11(11/6) + 10

y = -0.08333

The vertex is thus located at the point;

( 1.8333, -0.08333)

Find the attached

Part B:

We can use the quadratic formula to solve for x as follows;

The quadratic formula is given as,

$x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

From the quadratic equation given;

a = 3, b = -11, c = 10

We substitute these values into the above formula and simplify to determine the value of x;

$x=\frac{11+/-\sqrt{11^{2}-4(3)(10) } }{2(3)}=\frac{11+/-\sqrt{1} }{6}\\\\x=\frac{11+/-1}{6}\\\\x=\frac{11+1}{6}=2\\\\x=\frac{11-1}{6}=\frac{5}{3}$