Please help. Medal and fan :)
f(x) = -16x2 + 22x + 3
Part A: What are the x-intercepts of the graph of the f(x)? Show your work.
Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work.
Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.
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OpenStudy (anonymous):
@Jordan100 @bohotness @cwrw238
OpenStudy (anonymous):
@shortycme
OpenStudy (helder_edwin):
for part A set f(x)=0 and solve the quadratic equation.
for part B complete the square
OpenStudy (helder_edwin):
do u understand?
OpenStudy (anonymous):
not really :(
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OpenStudy (anonymous):
@helder_edwin
OpenStudy (helder_edwin):
sorry. a little bit busy.
OpenStudy (helder_edwin):
for part A set f(x)=0 so
\[ 0= f(x)=-16x^2+22+3\]
OpenStudy (helder_edwin):
can u solve this?
OpenStudy (anonymous):
im not good at math....
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OpenStudy (helder_edwin):
:-)
OpenStudy (helder_edwin):
ok the equation is
\[ 0=-16x^2+22x+3 \]
so we can use the quadratic formula.
OpenStudy (helder_edwin):
so
\[ x=\frac{-22\pm\sqrt{22^2-4(-16)3}}{2(-16)} \]
OpenStudy (anonymous):
thats part a right?
OpenStudy (helder_edwin):
yes
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
b and C? :)
OpenStudy (helder_edwin):
for part B u have to complete the square.
OpenStudy (helder_edwin):
do u know how to?
OpenStudy (helder_edwin):
to finish part A:
\[ x=\frac{-22\pm26}{-32} \]
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OpenStudy (helder_edwin):
we have to values:
\[ x_1=\frac{-22+26}{-32}=\frac{4}{-32}=-\frac{1}{8} \]
OpenStudy (helder_edwin):
and
\[ x_2=\frac{-22-26}{-32}=\frac{-48}{-32}=\frac{3}{2} \]