Yes, because there are common factors in the equation. therefore showing you can group
Answer:
Part A:
( 1.8333, -0.08333)
Part B:
x = 2 or x = 5/3
Step-by-step explanation:
The quadratic equation
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](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%283x%5E%7B2%7D-11x%2B10%29%3D6x-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}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%2B%2F-%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D)
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}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B11%2B%2F-%5Csqrt%7B11%5E%7B2%7D-4%283%29%2810%29%20%7D%20%7D%7B2%283%29%7D%3D%5Cfrac%7B11%2B%2F-%5Csqrt%7B1%7D%20%7D%7B6%7D%5C%5C%5C%5Cx%3D%5Cfrac%7B11%2B%2F-1%7D%7B6%7D%5C%5C%5C%5Cx%3D%5Cfrac%7B11%2B1%7D%7B6%7D%3D2%5C%5C%5C%5Cx%3D%5Cfrac%7B11-1%7D%7B6%7D%3D%5Cfrac%7B5%7D%7B3%7D)
Answer:
AAS
Step-by-step explanation: