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Finding Intercepts, Domain, Range and Vertex of a Parabola  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Finding information about a quadratic equation. So whenever we're trying to find a x intercept of a quadratic equation or any equation for that matter, what we do is let our y equal 0, so this is going to give us 0 is equal to x² plus 4x minus 3.
So all we have to do is find the zeros of this equation. We can either use factoring or quadratic formula. What we have here though is our coefficient when x is negative which is sort of how to deal with, so what we can actually do is bring everything to the other side if we want to or factor out a 1.
I'm just going to factor out a 1, so what we end up with is x² minus 4x plus 3. Now we just want to factor that down and we end up with x minus 3, x minus 1 giving us zeros of 1 and 3. So pretty easy way to deal with a negative sign is just to factor out.
So we found our x intercepts. The next thing we want to find is our y intercepts which we find by letting x equals 0. When we let x equals 0, our first two terms disappear and just leave us with 3, the last term.
To find the vertex, we can use b over 2a. Our b is 4, so we end up with 4 over our a is 1, so we end up with 2 which gives us 2. That gives us the x coordinate of the vertex, so in order to find the y coordinate, we just have to go and plug 2 back in.
So we plugged 2 back in we end up with 4 plus 8 minus 3 which gives us 4 minus 3, 1. So using b over 2a plug in to get the y coordinate, domain we're dealing with a parabola, our domain is always going to be all reals and our range, our range is our y values.
So for this one what we want to look at is our y coordinate the vertex which is 1. The negative on the x² tells us our parabola is facing downwards so we have a y coordinate of our vertex at 1 and our parabola goes downwards, sorry the y coordinate of the vertex at 1, parabola facing downwards, so we know that 1 is going to be the largest y value get. So our range then is just going to be from negative infinity up until 1 that point is on the curve so we include one with a hard bracket.
So finding a bunch of information about a quadratic intercepts just plug in 0 for the respective thing b over 2a for your vertex domain and range.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Finding Intercepts, Domain, Range and Vertex of a Parabola
Problem 1 8,412 viewsGiven f(x) = x² + 3x − 10
xint:yint:vertex:domain:range: 
Finding Intercepts, Domain, Range and Vertex of a Parabola
Problem 2 6,278 viewsGiven f(x) = x² + 4x − 3
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Problem 3 541 views 
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