SOLUTION: Graph the quadratic function f(x)=7(x+1)^2+4: state the ordeed pairs representing the x-intercept and y-intercept (if any) and the vertex or the parabola.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Graph the quadratic function f(x)=7(x+1)^2+4: state the ordeed pairs representing the x-intercept and y-intercept (if any) and the vertex or the parabola.      Log On


   

Question 369244: Graph the quadratic function f(x)=7(x+1)^2+4: state the ordeed pairs representing the x-intercept and y-intercept (if any) and the vertex or the parabola.
Found 2 solutions by stanbon, lefty4ever26:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Graph the quadratic function f(x)=7(x+1)^2+4:
state the ordeed pairs representing the x-intercept and y-intercept (if any) and the vertex or the parabola.
----
Vertex: (-1,4)
y-int: (0,11)
x-int: none
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graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C7%28x%2B1%29%5E2%2B4%29
=====
Cheers,
Stan H.


Answer by lefty4ever26(59) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=7%28x%2B1%29%5E2%2B4 I simplified it by FOILing the (x+1)^2 and multiplying 7 thru giving me
f%28x%29=7x%5E2%2B14x%2B11 The x-intercept occurs when f(x)=0.
0=7x%5E2%2B14x%2B11 Using the determinate sqrt%28b%5E2-4ac%29 plugging in 7,14,11 for a,b,c. you determine that There is no x intercept because you can't take the square root of a negative number.
The y intercept is found when x=0. So plug in 0 for any x.
7%280%29%5E2%2B14%280%29%2B11=11 SO the y intercept is at (0,11)
To find the vertex take the derivative of f(x) to get
f%27%28x%29=14x%2B14 set equal to 0 and solve for x.
0=14x%2B14
x=-1 Solve f(-1)
f%28-1%29=7-14%2B11=4 So your vertex is at (-1,4) which is illustrated in the following graph
graph%28300%2C200%2C-5%2C5%2C-1%2C15%2C7%28x%2B1%29%5E2%2B4%29