Question 3 (4 MARKS) Find the equation of the tangents to the curve of the function with rule y = x^2 at the points where x = a
1 answer:
Answer:
(0 , -a²)
Step-by-step explanation:
tangent at x = a and x = -a
y = x² (a , a²) and (-a , a²) must be on the curve and tangent to curve
gradient dy/dx = 2x
slope (m) at x = a is <u>2a</u> and slope (m') at x=-a is<u> -2a</u>
line1: (y - a²) / (x - a) = 2a
y - a² = 2a (x - a) y = 2ax - a² ... (1)
line2: (y - a²) / (x + a) = - 2a y = -2ax- a² ... (2)
(1) - (2): 4ax = 0 a≠0 x = 0
y = - a²
I wish I did it right, or ....
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well you didn't show problem 11 but here is a pretty photo of Port orford Oregon
<h3><u>Explanation</u></h3>
- Convert the equation into slope-intercept form.
where m = slope and b = y-intercept.
What we have to do is to make the y-term as the subject of equation.
- Simplify
From y = mx+b, the slope is 3.
<h3><u>Answer</u></h3>
To graph an equation with one x variable you first go on the y (vertical axis) and put a point at the number that doesnt have the x (12). Sometimes this number could be zero. The +12 is where it crosses the y axis. next to make the line you look at the number infront of x (1). that is the slope. It is basically 1 over 1. this means you go up one and right one, up one and right one, or down one and left one etc. this will make a line. If the number infront of x was -2 then it is seen as -2 over one. this slope is drawn by going down one and right one or up two and left one
