Answer:
[tex]\left(\dfrac{1}{4},\dfrac{49}{8}\right)[/tex]
Step-by-step explanation:
Turning points (stationary points) occur when the gradient of a graph is zero.
To find when the gradient of the graph is zero, differentiate the function, set it zero, then solve for x.
Given function:
[tex]f(x)=-2x^2+x+6[/tex]
Differentiate:
[tex]\implies f'(x)=(2)(-2)x^{2-1}+(1)x^{1-1}+6(0)[/tex]
[tex]\implies f'(x)=-4x+1[/tex]
Set the differentiated function to zero and solve for x:
[tex]\implies f'(x)=0[/tex]
[tex]\implies -4x+1=0[/tex]
[tex]\implies 4x=1[/tex]
[tex]\implies x=\dfrac{1}{4}[/tex]
To find the y-coordinate, input the found value of y into the given function:
[tex]\implies f\left(\dfrac{1}{4}\right)=-2\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{4}\right)+6[/tex]
[tex]\implies f\left(\dfrac{1}{4}\right)=\dfrac{49}{8}[/tex]
Therefore, the turning point of the function is:
[tex]\left(\dfrac{1}{4},\dfrac{49}{8}\right)[/tex]
answers
f=1/4=49/8
Step-by-step explanation:
identify the coefficients
a=-2, b=1
substitute the coefficient into the expression
x= -1/((2x(-2))
then solve it out the equation
x=1/4
evaluate the function x =1/4
f(x) =-2x²+x+6(1/4)
f=1/4=49/8
