What is the equation of the line to the parabola y=x^{2}+5x+1 through

sputneyoh

sputneyoh

Answered question

2021-12-10

What is the equation of the line to the parabola
y=x2+5x+1
through the point at
(12, 154)?

Answer & Explanation

Papilys3q

Papilys3q

Beginner2021-12-11Added 34 answers

Step 1
Given parabolic equation is y=x2+5x+1
Point (x1, y1)=(12, 154)
Differentiate the given equation
dydx=2x+5
(dydx)(12, 154)=2(12)+5=1+5=6
So, m=6
Step 2
Equation of line to the given parabola passing through
(x1, y1)=(12, 154)
with slope, m=6 is
yy1=m(xx1)
y154=6(x12)
4y154=6(2x12)
4y154=3(2x1)
4y15=12(2x1)
4y15=24x12
24x124y+15=0
24x4y+3=0
The required equation is 24x4y+3=0
Stella Calderon

Stella Calderon

Beginner2021-12-12Added 35 answers

Given:
y=x2+5x+1
point:
(12, 154)
1) dydx=2x+5
Using equation (1)
=2(12)+5
=1+5=6
with slope, m=6 is
2) yy1=m(xx1)
y154=6(x12)
4y154=6(2x12)
=3(2x1)
4y15=12(2x1)
=24x12
24x124y+15=0
Answer: 24x4y+3=0

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