# The function is given by the formula f(x)=2x^{2}-5x+3 a) find f(-1).b) at what values x, f(x)=1? c)Does the point A(1;0) belong to the graph of the function

The function is given by the formula $f\left(x\right)=2{x}^{2}-5x+3$
a) find f(-1).
b) at what values x, f(x)=1?
c)Does the point A(1;0) belong to the graph of the function
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Illuddybopylu
$a\right)f\left(-1\right)=2\ast \left(-1{\right)}^{2}-5\ast \left(-1\right)+3=2+5+3=10$
$b\right)2{x}^{2}-5x+3=1⇔2{x}^{2}-5x+2=0$
$D={5}^{2}-4\ast 2\ast 2=25-16=9$
x1=5+3/4=8/4=2
x2=5-3/4=2/4=0.5.
$c\right)2\ast {1}^{2}-5\ast 1+3=0$
2-5+3=0
-3+3=0
0=0
This point belongs to the graph.
###### Not exactly what you’re looking for?
mercuross8
$F\left(x\right)=2{x}^{2}-5x+3$
$1\right)F\left(-1\right)=2\ast \left(-1{\right)}^{2}-5\ast \left(-1\right)+3=2+5+3=10$
2)F(x)=1 find x
$2{x}^{2}-5x+3=1$
$2{x}^{2}-5x+2=0$
${x}^{2}-2.5x+1=0$
x1+x2=2.5
x1*x2=1
x1=0.5
x2=2
3)A(1.0)
x=1
$2\ast {1}^{2}-5\ast 1+3=0$