Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find the 
y-coordinate of the 
y-intercept of the polynomial function defined below.

f(x)=2x(3x+4)(2x+1)(2x-1)^(2)
Answer:

Find the y y -coordinate of the y y -intercept of the polynomial function defined below.\newlinef(x)=2x(3x+4)(2x+1)(2x1)2 f(x)=2 x(3 x+4)(2 x+1)(2 x-1)^{2} \newlineAnswer:

Full solution

Q. Find the y y -coordinate of the y y -intercept of the polynomial function defined below.\newlinef(x)=2x(3x+4)(2x+1)(2x1)2 f(x)=2 x(3 x+4)(2 x+1)(2 x-1)^{2} \newlineAnswer:
  1. Evaluate at x=0x = 0: To find the yy-coordinate of the yy-intercept of the function, we need to evaluate the function at x=0x = 0. This is because the yy-intercept occurs where the graph of the function crosses the yy-axis, and the xx-coordinate is always 00 at the yy-axis.
  2. Substitute x=0x = 0: Substitute x=0x = 0 into the function f(x)f(x) to find the y-coordinate of the y-intercept.\newlinef(x)=2x(3x+4)(2x+1)(2x1)2f(x) = 2x(3x+4)(2x+1)(2x-1)^{2}\newlinef(0)=20(30+4)(20+1)(201)2f(0) = 2\cdot 0\cdot (3\cdot 0+4)(2\cdot 0+1)(2\cdot 0-1)^{2}
  3. Simplify the expression: Simplify the expression by performing the multiplication and evaluating the powers.\newlinef(0)=20(0+4)(0+1)(01)2f(0) = 2\cdot0\cdot(0+4)(0+1)(0-1)^{2}\newlinef(0)=2041(1)2f(0) = 2\cdot0\cdot4\cdot1\cdot(-1)^{2}
  4. Continue simplifying: Continue simplifying the expression.\newlinef(0)=0×4×1×1f(0) = 0 \times 4 \times 1 \times 1\newlinef(0)=0f(0) = 0
  5. Final y-coordinate: The y-coordinate of the y-intercept is the value of f(0)f(0), which we have found to be 00.

More problems from Reflections of functions