Answer:
x=1
Step-by-step explanation:
According to your typed question,
f(x)=2x-3
Let f(x)=y
y=2x-3
Now,
Interchanging the positions of x and y
x=2y-3
x+3=2y
x+3/2=y
f'(x)=x+3/2
Then,
f(x)=f'(x)
2x-3=x+3/2
2(2x-3)=x+3
4x-3=x+3
4x-x=3+3
3x=3
x=3/3
x=1
According to your image question,
f(x)=x/2x-3
f(x)=f'(x)
Now,
Let y=f(x)
y=x/2x-3
y(2x-3)=x
2xy-3=x
2xy-x=3
x(2y-1)=3
2y-1=3x
2y=3x+1
y=3x+1/2
f'(x)=3x+1/2
Then,
f(x)=f'(x)
x/2x-3=3x+1/2
2x=6xsq+2x-9x+3
2x=6xsq+7x+3
solve for x ok
Are these 2 different questions or is this one equation
The simple/ <span>common sense method:
</span>The typical lay out of a quadratic equation is ax^2+bx+c
'c' represents where the line crosses the 'y' axis.
The equation is only translated in the 'y' (upwards/downwards) direction, therefore only the 'c' component of the equation is going to change.
A translation upwards of 10 units means that the line will cross the 'y' axis 10 places higher.
9+10=19,
therefore <u>c=19</u>.
The new equation is: <u>y=x^2+19 </u>
<span>
<span>The most complicated/thorough method:
</span></span>This is useful for when the graph is translated both along the 'y' axis and 'x' axis.
ax^2+bx+c
a=1, b=0, c=9
Find the vertex (the highest of lowest point) of f(x).
Use the -b/2a formula to find the 'x' coordinate of your vertex..
x= -0/2*1, your x coordinate is therefore 0.
substitute your x coordinate into your equation to find your y coordinate..
y= 0^2+0+9
y=9.
Your coordinates of your vertex f(x) are therefore <u>(0,9) </u>
The translation of upward 10 units means that the y coordinate of the vertex will increase by 10. The coordinates of the vertex g(x) are therefore:
<u>(0, 19) </u>
substitute your vertex's y coordinate into f(x)
19=x^2+c
19=0+c
c=19
therefore <u>g(x)=x^2+19</u>
Answer:
4
Step-by-step explanation:
