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Popular >

h(x)=log_{10}(2x^2+5x+2)

Solution

h (x) = lo g_{10} (2 x^{2} + 5 x + 2)

Solution

Domain : x < - 2 or x > - \frac{1}{2}

Range : - \infty < f (x) < \infty

X Intercepts : (\frac{- 5 + 17}{4}, 0), (\frac{- 5 - 17}{4}, 0), Y Intercepts : (0, lo g_{10} (2))

Asymptotes : Vertical x = - 2, x = - \frac{1}{2}

Extreme Points : None

+1

Interval Notation

Domain : (- \infty, - 2) \cup (- \frac{1}{2}, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

lo g_{10} (2 x^{2} + 5 x + 2) : x < - 2 or x > - \frac{1}{2}

Range of

lo g_{10} (2 x^{2} + 5 x + 2) : - \infty < f (x) < \infty

Axis interception points of

lo g_{10} (2 x^{2} + 5 x + 2) :

X Intercepts

: (\frac{- 5 + 17}{4}, 0), (\frac{- 5 - 17}{4}, 0),

Y Intercepts

: (0, lo g_{10} (2))

Asymptotes of

lo g_{10} (2 x^{2} + 5 x + 2) :

Vertical

: x = - 2, x = - \frac{1}{2}

Extreme Points of

lo g_{10} (2 x^{2} + 5 x + 2) :

None

Graph

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Frequently Asked Questions (FAQ)

What is the h(x)=log_{10}(2x^2+5x+2) ?
The h(x)=log_{10}(2x^2+5x+2) is