# Category Archives: Rock

1. Zulkigar says:
An inflection point is a point on a curve at which the sign of the curvature (i.e., the concavity) changes. Inflection points may be stationary points, but are not local maxima or local minima. For example, for the curve plotted above, the point is an inflection point.4/5(3).
2. Grokora says:
May 31,  · Let (x,y) be the point on the given curve which is closest to the point(3,0) where The distance between the point (x,y) and (3,0) is Now, we will find the critical points of f(x). For this value of x, Therefore, f(x) is minimum, that is distance between points (x,y) and (3,0) is minimum for x=5/2.
3. Nat says:
Oct 10,  · (-2,21) (1,6) step one: find the derivative of the equation. y'=6x^2+6x Step two: Since a horizontal line has a slope of 0, set the derivative to equal 0 and solve. y' = 6(x^2+x-2) y' = 6(x+2)(x-1) x= -2, 1 Step three: plug the x-values found in step 2 back into the original equation to get the y-coordinates of the points on the curve. y(-2)= 21 y(1)= -6 Step four: write out the coordinates.
4. Mazulkis says:
Find the point(s) on the curve $y^3=x^2$ closest to the point $P=(0,4).$ I understand that there is a way to solve this, using the distance formula, however this.
5. Zolojinn says:
Sep 27,  · Find the points on the curve? Find the points on the curve given below, where the tangent is decerkaltmasmortbemejaffvanvicharlai.xyzinfo to three decimal places: y = 9x^3 + 7x^2 - 3x + 5.
6. Grojinn says:
Jan 07,  · Transcript. Ex , 27 (Method 1) The point on the curve 𝑥2=2𝑦 which is nearest to the point (0, 5) is (A) (2 √2,4) (B) (2 √2,0) (C) (0, 0) (D) (2, 2) Let (ℎ, 𝑘) be the point on the curve 𝑥2 = 2𝑦 Where is nearest to the point (0, 5) Since (ℎ, 𝑘) lie on the curve 𝑥2= 2𝑦 ⇒ (ℎ 𝑘) will satisfy the equation of curve 𝑥2=2𝑦 ⇒ Putting 𝑥=ℎ & y=𝑘.
7. Mozil says:
Example 2: Find the equation of the tangent and the normal to the curve y = x 4 – 3x 3 + 6x + 2 at the point (2, 6) Step 1: First find the gradient of the tangent at the point (2, 6) y’ = = 4x 3 – 9x 2 + 6. at x = 2 = m T = 4 x 2 3 – 9 x 2 2 + 6. m T = 2 (m T is the gradient or slope of the tangent).
8. Nakora says:
Find the point where the tangent to the curve is horizontal. {eq}f(x) = \frac{x^x+4}{x^2+4} {/eq} The Horizontal Tangent: Let us assume that we have function {eq}y=f(x). {/eq} The tangent line.
9. Kashicage says:
We are trying to find the point A (x,y) on the graph of the parabola, y = x 2 + 1, that is closest to the point B (4,1). Restatement of the problem: Find the point A (x,y) on the graph of the parabola, y = x 2 + 1, that minimizes the distance d between the curve and the point B .