Question

1. the following 2 graphs intersect at a point. what is the x value of the intersection?

y = 3x² - 2x + 3 and -2x³ + 2x - 1
x = ________ (round to the nearest hundredth)

Explanation:

Step1: Set the equations equal

To find the intersection, set \(3x^2 - 2x + 3=-2x^3 + 2x - 1\). Rearrange to get \(2x^3+3x^2 - 4x + 4 = 0\).

Step2: Use numerical method (e.g., Newton - Raphson)

Let \(f(x)=2x^3 + 3x^2-4x + 4\). First, find \(f(-2)=2(-8)+3(4)-4(-2)+4=-16 + 12 + 8 + 4 = 8\), \(f(-3)=2(-27)+3(9)-4(-3)+4=-54 + 27+12 + 4=-11\). So a root is between \(-3\) and \(-2\).
Take \(x_0=-2.5\), \(f(-2.5)=2(-15.625)+3(6.25)-4(-2.5)+4=-31.25 + 18.75 + 10 + 4 = 1.5\)
\(f(-2.6)=2(-17.576)+3(6.76)-4(-2.6)+4=-35.152+20.28 + 10.4+4=-0.472\)
Now, use linear approximation between \(x=-2.6\) (\(f=-0.472\)) and \(x=-2.5\) (\(f = 1.5\)). The difference in \(x\) is \(0.1\), difference in \(f\) is \(1.5-(-0.472)=1.972\). We need to find \(\Delta x\) such that \(0 - (-0.472)=\frac{1.972}{0.1}\Delta x\), \(\Delta x=\frac{0.472\times0.1}{1.972}\approx0.024\). So \(x\approx - 2.6+0.024=-2.576\approx - 2.58\) (checking \(f(-2.58)=2(-2.58)^3+3(-2.58)^2-4(-2.58)+4\approx2(-17.174)+3(6.656)+10.32 + 4\approx - 34.348+19.968+10.32 + 4\approx - 0.06\)), \(f(-2.57)=2(-17.003)+3(6.605)+10.28 + 4\approx - 34.006+19.815+10.28 + 4\approx0.089\). Using linear approximation between \(x=-2.58\) (\(f=-0.06\)) and \(x=-2.57\) (\(f = 0.089\)): \(\Delta x=\frac{0 - (-0.06)}{0.089-(-0.06)}\times0.01\approx\frac{0.06}{0.149}\times0.01\approx0.004\). So \(x\approx - 2.58 + 0.004=-2.576\approx - 2.58\) (more accurately, using a calculator or better method, the real root is approximately \(-2.58\)).

Answer:

\(-2.58\)