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Math  /  Algebra

QuestionFind the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necessary). y=−2x2+10x+145y=-2 x^{2}+10 x+145

Studdy Solution

STEP 1

What is this asking? We're looking for the *x*-values where this parabola crosses the *x*-axis (the roots!), and the tip of the parabola, which is called the vertex! Watch out! Remember, the quadratic formula can be your best friend here, but make sure you plug in the values correctly!
Also, don't forget that a parabola can have two roots, one root, or no roots at all!

STEP 2

1. Find the roots.
2. Find the vertex.

STEP 3

Let's **kick things off** with the quadratic formula!
It's like a magic key for unlocking those roots.
Remember, the *roots* are where our parabola y=−2x2+10x+145y = -2x^2 + 10x + 145 crosses the *x*-axis, which means y=0y = 0.
So we're solving −2x2+10x+145=0-2x^2 + 10x + 145 = 0.

STEP 4

The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
In our equation, a=−2a = \mathbf{-2}, b=10b = \mathbf{10}, and c=145c = \mathbf{145}.
Let's **plug these values** in!

STEP 5

x=−(10)±(10)2−4(−2)(145)2(−2)x = \frac{-(\mathbf{10}) \pm \sqrt{(\mathbf{10})^2 - 4(\mathbf{-2})(\mathbf{145})}}{2(\mathbf{-2})}

STEP 6

x=−10±100+1160−4x = \frac{-10 \pm \sqrt{100 + 1160}}{-4}

STEP 7

x=−10±1260−4x = \frac{-10 \pm \sqrt{1260}}{-4}

STEP 8

Now, let's **crank out** those two possible *x* values: x=−10+1260−4≈−10+35.496−4≈25.496−4≈−6.374x = \frac{-10 + \sqrt{1260}}{-4} \approx \frac{-10 + 35.496}{-4} \approx \frac{25.496}{-4} \approx \mathbf{-6.374} and x=−10−1260−4≈−10−35.496−4≈−45.496−4≈11.374x = \frac{-10 - \sqrt{1260}}{-4} \approx \frac{-10 - 35.496}{-4} \approx \frac{-45.496}{-4} \approx \mathbf{11.374} So, our roots are approximately x=−6.374x = \mathbf{-6.374} and x=11.374x = \mathbf{11.374}!

STEP 9

The *x*-coordinate of the vertex is given by x=−b2ax = \frac{-b}{2a}.
Using our **values** for *a* and *b*, we get:

STEP 10

x=−102(−2)=−10−4=2.5x = \frac{-10}{2(-2)} = \frac{-10}{-4} = \mathbf{2.5}

STEP 11

Now, let's **plug this *x* value** back into our original equation to find the *y*-coordinate of the vertex: y=−2(2.5)2+10(2.5)+145y = -2(\mathbf{2.5})^2 + 10(\mathbf{2.5}) + 145

STEP 12

y=−2(6.25)+25+145y = -2(6.25) + 25 + 145

STEP 13

y=−12.5+25+145y = -12.5 + 25 + 145

STEP 14

y=157.5y = \mathbf{157.5}

STEP 15

So, our vertex is at (2.5,157.5)(\mathbf{2.5}, \mathbf{157.5})!

STEP 16

The roots of the quadratic are approximately x=−6.374x = \mathbf{-6.374} and x=11.374x = \mathbf{11.374}.
The vertex is at (2.5,157.5)(\mathbf{2.5}, \mathbf{157.5}).

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