Step 1: convert the equation into fhe vertex form
To do you can complete squares:
y = - [x^2 + 4x + 3]
y = - [ (x + 2)^2 - 4 + 3]
y = - [ (x + 2)^2 - 1] = - (x+2)^2 + 1
Then the vertex is (-2, 1)
Now you can drasw the vertex
Step 2: Find the roots (zeros)
y = - [ (x + 2)^2 - 1] = 0
(x + 2)^2 - 1 = 0
(x+2)^2 = 1
(x+2) = (+/-) √1
x + 2 = (+/-1)
x = - 2 +/- 1
x = -1 and x = -3
Now you draw the points (-1,0) , (-3,0)
Step 3: find the interception with the y-axis.
That is y value when x = 0
y = - (0)^2 - 4(0) - 3 = -3
Then draw the point (0, -3)
Step 4: given that the coefficent of x is negative (-1) the parabola is open downward.
So, with those four points: vertex (-2,1), (-1,0), (-3,0) and (0,-3), you can sketch the function.
The equation of the line is y=6x+37
Step-by-step explanation:
You can write the equation using the slope intercept form techniques which is y=mx+c where m is the slope of the line and cis they intercept constant.
Given the point (-6,1) and slope m=6then
Δy/Δx =m
![y-1/x--6 = 6\\\\y-1/x+6=6\\\\\\y-1=6(x+6)\\\\\\y-1=6x+36\\\\\\y=6x+36+1\\\\y=6x+37](https://tex.z-dn.net/?f=y-1%2Fx--6%20%3D%206%5C%5C%5C%5Cy-1%2Fx%2B6%3D6%5C%5C%5C%5C%5C%5Cy-1%3D6%28x%2B6%29%5C%5C%5C%5C%5C%5Cy-1%3D6x%2B36%5C%5C%5C%5C%5C%5Cy%3D6x%2B36%2B1%5C%5C%5C%5Cy%3D6x%2B37)
If you were to graph y=x it would look like if you took the x axis and rotated it 45 degrees counterclockwise with the center of rotation on the origin
basically, it switches x and y
a reflection is like a miror
a reflection across y=x in equation would be
y=3x^2+5x-2 reflected across the y=x axis (aka the identity line) would be
x=3y^2+5y-2
<span>7n-2(n+5)< 3n-16 / -3n-16
7n-2(n+5)-3n+16<0
5n-3n-10+16<0
2n-10+16<0
2n+6<0 / -6
2n< -6 / :2
n< -6/2
n< -3
n</span>є(-∞:-3)
Answer:
![y =\dfrac{12}{5}x + 22](https://tex.z-dn.net/?f=y%20%3D%5Cdfrac%7B12%7D%7B5%7Dx%20%2B%2022)
Step-by-step explanation:
Slope-intercept form of line:
First find the slope of the line AB. ie, m
Slope of the perpendicular line = -1/m
(2 , 3) ⇒ x₁ = 2 & y₁ = 3
(-10, 8) ⇒ x₂ = -10 & y₂ = 8
![\boxed{Slope=\dfrac{y_2-y_1}{x_2-x_1}}](https://tex.z-dn.net/?f=%5Cboxed%7BSlope%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%7D)
![= \dfrac{8-3}{-10-2}\\\\=\dfrac{5}{-12}\\\\=\dfrac{-5}{12}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B8-3%7D%7B-10-2%7D%5C%5C%5C%5C%3D%5Cdfrac%7B5%7D%7B-12%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-5%7D%7B12%7D)
![\sf slope \ of \ the \ perpendicular \ line \ m_1 = \dfrac{-1}{m}= -1 \ \div \dfrac{-5}{12}](https://tex.z-dn.net/?f=%5Csf%20slope%20%5C%20of%20%5C%20the%20%5C%20perpendicular%20%5C%20line%20%5C%20m_1%20%3D%20%5Cdfrac%7B-1%7D%7Bm%7D%3D%20-1%20%5C%20%5Cdiv%20%5Cdfrac%7B-5%7D%7B12%7D)
![\sf = -1 * \dfrac{12}{-5}=\dfrac{12}{5}](https://tex.z-dn.net/?f=%5Csf%20%3D%20-1%20%2A%20%5Cdfrac%7B12%7D%7B-5%7D%3D%5Cdfrac%7B12%7D%7B5%7D)
Equation of the required line: y = mx + b
![y =\dfrac{12}{5}x+b](https://tex.z-dn.net/?f=y%20%3D%5Cdfrac%7B12%7D%7B5%7Dx%2Bb)
The line passes through (-5 , 10). Substitute in the above equaiton,
![10 =\dfrac{12}{5}*(-5) + b\\\\ 10 = (-12) + b\\\\](https://tex.z-dn.net/?f=10%20%3D%5Cdfrac%7B12%7D%7B5%7D%2A%28-5%29%20%2B%20b%5C%5C%5C%5C%2010%20%3D%20%28-12%29%20%2B%20b%5C%5C%5C%5C)
10 + 12 = b
b = 22
Equation of the line:
![y =\dfrac{12}{5}x + 22](https://tex.z-dn.net/?f=y%20%3D%5Cdfrac%7B12%7D%7B5%7Dx%20%2B%2022)