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Math  /  Algebra

Questionf(x)=−3x2+2x+3f(x)=-3 x^{2}+2 x+3
Round to the nearest hundredth if necessary. If there is more than one xx-intercept, separate them If applicable, click on "None". \begin{tabular}{|ll|} \hline vertex: & (II, □\square \\ xx-intercept(s): & □\square \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. The function given is a quadratic function of the form f(x)=ax2+bx+c f(x) = ax^2 + bx + c .
2. The vertex of a quadratic function in standard form can be found using the formula x=−b2a x = -\frac{b}{2a} .
3. The x x -intercepts (roots) of the quadratic function can be found using the quadratic formula x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .

STEP 2

1. Identify the coefficients a a , b b , and c c .
2. Calculate the vertex of the quadratic function.
3. Calculate the x x -intercepts using the quadratic formula.

STEP 3

Identify the coefficients from the quadratic function f(x)=−3x2+2x+3 f(x) = -3x^2 + 2x + 3 .
Here, a=−3 a = -3 , b=2 b = 2 , and c=3 c = 3 .

STEP 4

Calculate the x x -coordinate of the vertex using the formula x=−b2a x = -\frac{b}{2a} .
x=−22(−3)=−2−6=13 x = -\frac{2}{2(-3)} = \frac{-2}{-6} = \frac{1}{3}

STEP 5

Substitute x=13 x = \frac{1}{3} into the function to find the y y -coordinate of the vertex.
f(13)=−3(13)2+2(13)+3 f\left(\frac{1}{3}\right) = -3\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) + 3
=−3(19)+23+3 = -3\left(\frac{1}{9}\right) + \frac{2}{3} + 3
=−13+23+3 = -\frac{1}{3} + \frac{2}{3} + 3
=13+3 = \frac{1}{3} + 3
=13+93 = \frac{1}{3} + \frac{9}{3}
=103 = \frac{10}{3}
The vertex is (13,103) \left(\frac{1}{3}, \frac{10}{3}\right) .

STEP 6

Calculate the x x -intercepts using the quadratic formula x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .
First, calculate the discriminant:
b2−4ac=22−4(−3)(3) b^2 - 4ac = 2^2 - 4(-3)(3)
=4+36 = 4 + 36
=40 = 40

STEP 7

Substitute the discriminant into the quadratic formula:
x=−2±402(−3) x = \frac{-2 \pm \sqrt{40}}{2(-3)}
=−2±40−6 = \frac{-2 \pm \sqrt{40}}{-6}
Simplify 40 \sqrt{40} :
40=4×10=210 \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}
x=−2±210−6 x = \frac{-2 \pm 2\sqrt{10}}{-6}
=−2−6±210−6 = \frac{-2}{-6} \pm \frac{2\sqrt{10}}{-6}
=13∓103 = \frac{1}{3} \mp \frac{\sqrt{10}}{3}
The x x -intercepts are:
x=1+103andx=1−103 x = \frac{1 + \sqrt{10}}{3} \quad \text{and} \quad x = \frac{1 - \sqrt{10}}{3}
Calculate these values to the nearest hundredth:
x≈1+3.163≈4.163≈1.39 x \approx \frac{1 + 3.16}{3} \approx \frac{4.16}{3} \approx 1.39
x≈1−3.163≈−2.163≈−0.72 x \approx \frac{1 - 3.16}{3} \approx \frac{-2.16}{3} \approx -0.72
The vertex is (13,103)≈(0.33,3.33) \left(\frac{1}{3}, \frac{10}{3}\right) \approx (0.33, 3.33) .
The x x -intercepts are approximately 1.39 1.39 and −0.72 -0.72 .

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