Question

Y=2x^2-32
Find the x-intercepts of this equation how do u do that ?

Answer 1

Help some one :(

Answer 2

x-intercepts of a function occur at values of x that cause y to equal zero. To find x-intercepts, substitute 0 for y and solve for x.

\[0=2x ^{2}-32 \]
\[32=2x ^{2}\] (Add 32 to both sides)
\[x^2=16 \] (divide both sides by 2)
\[\sqrt{x^2}=\pm \sqrt{16}\] (Take the square root of both sides)
\[x=\pm4\]

Therefore, the x-intercepts of the function are 4 and -4. If you graph the function, you should see the curve intersect the x-axis at those values. If you substitute 4 or -4 in for x, you should get 0.