SOLUTION: We are given the following equation: f(x)=x^2+25, and are asked to find the Vertex and the X-intercept(s).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: We are given the following equation: f(x)=x^2+25, and are asked to find the Vertex and the X-intercept(s).      Log On


   

Question 1128688: We are given the following equation: f(x)=x^2+25, and are asked to find the Vertex and the X-intercept(s).
Found 2 solutions by josmiceli, MathLover1:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+x%5E2+%2B+25+
The x-intercept(s) is where +y=0+, so
+x%5E2+%2B+25+=+0+
+x%5E2+=+-25+
+x+=+%2Bsqrt%28+-25+%29+
+x+=+5i+
and, also,
+x+=+-5i+
-----------------
There are 2 imaginary x-intercepts, and no
real x-intercepts
-----------------
Using the general form:
+y+=+a%2Ax%5E2+%2B+b%2Ax+%2B+c+
The x-value of the vertex is at:
+x%5Bv%5D+=+-b%2F%282a%29+ where
+a+=+1+
+b+=+0+
+x%5Bv%5D+=+0%2F%282%2A1%29+
+x%5Bv%5D+=+0+
Plug this back into equation:
+y+=+0%5E2+%2B+25+
The vertex is at ( 0,25 )
--------------------------------
Here's the plot:
+graph%28+400%2C+400%2C+-10%2C+10%2C+-5%2C+100%2C+x%5E2+%2B+25+%29+

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=x%5E2%2B25, => parabola, opening up, has minimum
to find the Vertex, find minimum; set x=0
f%28x%29=0%5E2+%2B+25+
f%28x%29=25+

minimum or vertex is at (0,25)

and to find the x-intercept(s), set f%28x%29=0
0=x%5E2%2B25
-25=x%5E2
x=sqrt%28-25%29
x=-5i or x=5i=> there is no x-intercept(s)


+graph%28600%2C+600%2C+-30%2C+30%2C+-30%2C+30%2C+x%5E2%2B25%29+